The graph of g(x) is the graph of f(x) translated 2 units left by the operation g(x)=f(x+2) so option (D) is correct.
What is the transformation of a graph?Transformation is rearranging a graph by a given rule it could be either increment of coordinate or decrement or reflection.
If we reflect any graph about y = x then the coordinate will interchange it that (x,y) → (y,x).
If a function f(x) is transformed by funciton g(x) as shown,
g(x) = f(x+a)
For a>0, then the graph of f(x) shifts left by "a" unit, while if a<0, then the graph of f(x) shifts right side by "a"units.
As per the given function,
g(x) = f(x + 2)
Since 2 > 0 therefore the function will shift 2 units left.
Hence "The graph of g(x) is the graph of f(x) translated 2 units left by the operation g(x)=f(x+2)".
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Let f(x)=5x.Let g(x)=5x−7.Which statement describes the graph of g(x)with respect to the graph of f(x)? g(x)is translated 7 units down fromf(x).g(x)is translated 7 units left fromf(x).g(x)is translated 7 units right from f(x).g(x)is translated 7 units up fromf(x).
Given
[tex]\begin{gathered} f(x)=5x \\ g(x)=5x-7 \end{gathered}[/tex]According to rules of transformation:
f(x)+c shift c units up and f(x)-c shift c units down.
For the given function g(x) = 5x-7, 7 is being subtracted from 5x.
Where 5x is represented by f function.
Therefore, we could apply the rules of transformation f(x)-c shift c units down.
Here the value of c is 7.
Answer: g(x) is translated 7 units down from f(x)
In the rhombus m<1 = 160 what are m<2 and m<3. This diagram is not drawn to scale. Show all work
We are given a rhombus shape.
The measure of angle ∠1 = 160°
Recall that in a rhombus, the oppsite angles are equal, this means ∠1 = ∠2
So, ∠2 = 160°
Recall that the sum of all four interior angles in a rhombus must be equal to 360°
The diagonal line divides the angles in half.
This means that angle 3 and angle x are equal.
[tex]\begin{gathered} 160\degree+160\degree+2(\angle3+x)=360\degree_{} \\ 320\degree+2(\angle3+x)=360\degree \\ 2(\angle3+x)=360\degree-320\degree \\ 2(\angle3+x)=40\degree \\ \angle3+x=\frac{40\degree}{2} \\ \angle3+x=20\degree \end{gathered}[/tex]Since we know that ∠3 and ∠x are equal then
∠3 = 10° and ∠x = 10°
Therefore,
∠2 = 160°
∠3 = 10°
A class had a quiz where scores ranged from 0 to 10.N(s) models the number of students whose score on the quiz was s.What does the statement N(8) > N(5) mean?Group of answer choicesA score of 8 is greater than a score of 5.There are more students who scored 8 than students who scored 5.There are 8 students who scored higher than 5.
The expression N(s) models the number of students that got a score "s" on the quiz.
Then the expression N(8) represents the number of students that scored 8 on the quiz.
And N(5) represents the number of students that scored 5 on the quiz.
[tex]N(8)>N(5)[/tex]Can be read as: "The number of students whose score on the quiz was 8 is greater than the number of students whose score on the quiz was 5"
Therefore, you can conclude that there were more students who scored 8 than students who scored 5. (option 2)
In New York, the tax on a property assessed at $520,000 is $10,400. If tax rates are proportional in this city, how much would the tax be on a property assessed at $370,000? Answer: $
Given that the tax on a property assessed at $520,000 is $10,400 and the tax
Use a calculator to evaluate the expression. (Do not round until the final answer. Then round to three decimal places as needed.)
2.303
1) For the following expression:
[tex]\frac{\ln30+\ln15}{\log_{10}30+\log_{10}15}[/tex]We can simplify that and then round it off to the nearest thousandth:
2) Let's rewrite them simplifying using the logarithm property of multiplication:
[tex]\begin{gathered} \frac{\ln30+\ln15}{\log_{10}30+\log_{10}15}= \\ \frac{\ln(30\cdot15)}{\log_{10}30+\log_{10}15}= \\ \frac{\ln(30\cdot15)}{\log_{10}(30\cdot15)}= \\ \frac{\ln(450)}{\log_{10}(450)}= \end{gathered}[/tex]Note that the base of the Natural Log is the Euler's number "e" so let's move on now using the calculator, finally:
[tex]\frac{\ln(450)}{\log_{10}(450)}=\frac{6.10924}{2.65321}=2.30258\ldots\approx2.303[/tex]Note that only at the last step we have rounded it off. And that's the
answer
What is the missing coefficient of the x-term of the product (−x−5)^2 after it has been simplified?−25−101025
Given:
The terms is
[tex](-x-5)^2[/tex]Required:
What is the missing coefficient of the x-term of the product after it has been simplified?
Explanation:
We have to find the missing coefficient of the x term of the given product
We know
[tex](a-b)^2=a^2-2ab+b^2[/tex]So,
[tex](-x-5)^2=x^2+10x+25[/tex]Therefore, the missing coefficient of the x-term is 10.
Answer:
Therefore, the missing coefficient of the x-term is 10.
What 3D shape will be formed when the following are rotated around the axis
a)
A washer will be formed
b)
A cone will be formed
C)
A sphere will be formed
ava's family drove to disneyland for spring break. Her mom and dad shared the driving duties for a total of 24 hours. Her mom drove 75 miles per hour, and her dad drove 60 miles per hour. If they drove a total of 1,710 miles, how many hours did each person drive for?
Total driving time =24
Mom drove =75 mile per hours
Dad drove = 60 miles per hours
Total distance =1710
Let
[tex]\begin{gathered} \text{ mom driving time =}^{}t_1 \\ \text{dad driviving time=}^{}t_2 \\ \text{Mom driving distance =}x \\ \text{ So dad driving distance=}^{}1710-x \end{gathered}[/tex]Total time:
[tex]t_1+t_2=24[/tex]Formula:
[tex]\text{ Spe}ed=\frac{\text{ Distance}}{\text{ Time}}[/tex]For Ava's mom:
[tex]\begin{gathered} \text{Speed}=\frac{\text{ Distance}}{\text{ time}} \\ 75=\frac{x}{t_1} \\ x=75t_1^{} \end{gathered}[/tex]For Ava's dad:
[tex]\begin{gathered} \text{ Spe}ed=\frac{\text{ Distance}}{\text{ Time}} \\ 60=\frac{1710-x}{t_2} \\ 60t_2=1710-x \\ x=1710-60t_2 \end{gathered}[/tex]Put the value of "x" then:
[tex]\begin{gathered} x=75t_1 \\ x=1710-60t_2 \\ so\colon \\ 75t_1=1710-60t_2 \\ 75t_1+60t_2=1710 \\ 15(5t_1+4t_2)=15\times114 \\ 5t_1+4t_2=114 \end{gathered}[/tex]Solve the both eq then:
[tex]\begin{gathered} t_1+t_2=24 \\ 4t_1+4t_2=96 \\ 5t_1+4t_2=114 \\ \text{then:} \\ 5t_1-4t_1+4t_2-4t_2=114-96 \\ t_1=18 \\ \end{gathered}[/tex]So Ava's mom drive 18 hours
[tex]\begin{gathered} t_1+t_2=24 \\ 18+t_2=24 \\ t_2=24-18 \\ t_2=6 \end{gathered}[/tex]Ava's dad driving 6 houras
Soue se compound inequality and give your answer in intentel notation- 10 AND-80-72-1
S= (-4, 1 ]
1) Solving that compounded inequality
4x +6 > -10 and -8x+7 ≥ -1
2) Let's start by 4x +6 > -10
4x +6 > -10 Subtracting 6 from both sides
4x > -10-6
4x > -16 Dividing both sides by 4
x > -4
And with -8x+7 ≥ -1
-8x+7 ≥ -1 Subtract 7 from both sides
-8x ≥ -1 -7
-8x ≥ -8 Multiply by -1
8x ≤ 8
x ≤ 1
3) Graphing the solution interval:
So the solution is the interval S= (-4, 1 ] not including x= -4 and including the value x = 1
Melina made a scale drawing of a building.She used a scale in which 0.5 inch represents 1 foot. Which graph represents this relationship?
From the graph, the y - axis 10 uints while the x - axis is 5 units
The x - axis is labeled inches and its half of the feet
For every half inch on x - axis you have 1 feet
The graph that displays the scale is graph D
The answer is OPTION D
2) Add or subtract the following polynomials: (5pts each) 1) (98-7x' +5x-3)+(2x* +4x'-6x-8) = ii) (8x* +6x - 4x2 -2)-(3x* – 5x – 7x+9)=
When we are adding/subtracting polynomials, we add or subtract like terms.
For example,
x^2 added with x^2 terms
x^4 added with x^4 terms
numbers (constants) added with numbers etc.
2 i)[tex](9x^5-7x^2+5x-3)+(2x^4+4x^3-6x-8)[/tex]Since we are "adding" the 2nd parenthesis polynomial, we can take out the parenthesis and put them in order and them simply add/subtract(!) The steps are shown below:
[tex]\begin{gathered} (9x^5-7x^2+5x-3)+(2x^4+4x^3-6x-8) \\ =9x^5-7x^2+5x-3+2x^4+4x^3-6x-8 \\ =9x^5+2x^4+4x^3-7x^2+5x-6x-3-8 \\ =9x^5+2x^4+4x^3-7x^2-x-11 \end{gathered}[/tex]Note: there were like terms with "x's" and "constants". We added/subtracted them only.
Solve the inequality X - (5 - 3x) = 2x - 1
hi,
x - (5 - 3x) = 2x - 1
x - 5 + 3x = 2x - 1
x - 2x + 3x = -1 + 5
2x = 4
x = 4/2
x =< 2
[tex]\text{ x }\leq\text{ 2}[/tex]
The result is letter A, the first choice
can somone hep me please
Hi
a) = (8x2) x (10 ‐³ x10 ‐⁴)
= 8 x 2 you get 16 then 10‐³-⁴
16 x 10 ‐⁷
= 1.6 x 10¹ x 10 ‐⁷
= 1.6 x 10 ‐⁶
final answer
1.6 x 10 ‐⁶
Solve the equation-3 + a = 13a = ???
ANSWER
a = 16
EXPLANATION
To solve for a we have to add 3 on both sides of the equation:
[tex]\begin{gathered} -3+3+a=13+3 \\ a=16 \end{gathered}[/tex]At the airport, the new runway will be parallel to a nearby highway. The equation that represents the highway is 6y = 8x - 11. Which equation could represent the new runway? A. 9y = 12x + 5B. 9x = 12y + 8C. 12y = -9x + 2 D. 12x = -9y + 4
At the airport, the new runway will be parallel to a nearby highway. The equation that represents the highway is 6y = 8x - 11. Which equation could represent the new runway?
A. 9y = 12x + 5
B. 9x = 12y + 8
C. 12y = -9x + 2
D. 12x = -9y + 4
______________________________________________________
Parallel equations have the same slope
6y = 8x - 11
y= 8/6 x - 11/6
y= 4/3 x-11/6
y = m x +b (m is the slope )
_____________________________________________
You need to find the other equation with the same slope
____________________________
A. 9y = 12x + 5
y = 12/ 9 x + 5/9
y = 4/3 x + 5/9
_________________________
B. 9x = 12y + 8
12 y= 9x-8
y= 9/ 12 x- 8/12
y= 3/4 x - 4/6
discarded
________________________
C. 12y = -9x + 2
y = -9/12 x + 2/12
y = -3/4 x + 1/6
discarded
_________________________
D. 12x = -9y + 4
12x -4 = -9y
y = -12/9 x +4/9
y = -4/3 x+4/9
discarded
_______________
So then, A 9y = 12x + 5 is the equation that could represent the new runway because is parallel to the highway 6y = 8x - 11.
Two planes, which are 2320 miles apart, fly toward each other. Their speeds differ by 80 mph. If they pass each other in 4 hours,what is the speed of each?Step 1 of 2: Use the variable x to set up an equation to solve the given problem. Set up the equation, but do not take steps to solve it.
Given the word problem, we can deduce the following information.
1. Two planes, which are 2320 miles apart, fly toward each other.
2. Their speeds differ by 80 mph.
3. They pass each other in 4 hours.
To find the speed of each plane, we use the formula:
distance = (rate)(time)
Since they are flying towards each other, the sum of both speeds is 2x+80. So,
distance = (rate)(time)
2320 miles = (2x+80 mph)(4 hrs)
Thus, the equation to solve this is:
2320 = (2x+80)(4)
Jina spends $16 each time she travels the toll roads. She started the month with $240 in her toll road account. The amount, A (in dollars), that she has left in the account after t trips on the toll roads is given by the following function.=A(t)=240-16tAnswer the following questions.(a)How much money does Jina have left in the account after 11 trips on the toll roads?$(b)How many trips on the toll roads can she take until her account is empty?trips
GIVEN:
We are told that Jina had an opening balance of $240 in her toll road account.
Also, we are told that the amount left in the toll road account is given by the function;
[tex]A(t)=240-16t[/tex]Required;
(a) To find how much money she has left in her acount after 11 trips.
(b) To find out how many trips she can take until her account is empty.
Step-by-step solution;
We first take note of the variable t, which represent the number of trips taken. Also, the function shows how many trips multiplied by 16 would be subtracted from the opening balance. The result would be how much amount (variable A) would be left in her account.
Therefore;
(a) After 11 trips, Jina would have;
[tex]\begin{gathered} A(t)=240-16t \\ \\ A(11)=240-16(11) \\ \\ A(11)=240-176 \\ A(11)=64 \\ \end{gathered}[/tex]For the (A) part, the answer is $64.
(b) For her account to be empty, then the function given would be equal to zero. That is, after an unknown number of trips, the balance would be zero. We can now re-write the function as follows;
[tex]\begin{gathered} A(t)=240-16t \\ \\ 0=240-16t \end{gathered}[/tex]Add 16t to both sides of the equation;
[tex]\begin{gathered} 16t=240-16t+16t \\ \\ 16t=240 \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }16: \\ \\ \frac{16t}{16}=\frac{240}{16} \\ \\ t=15 \end{gathered}[/tex]This means after 15 trips she would have emptied her toll road account.
ANSWER:
[tex]\begin{gathered} (A)=\text{\$64} \\ \\ (B)=15\text{ }trips \end{gathered}[/tex]I need help finding the exact perimeter. Special right triangles.
Answer:
The exact perimeter of the square is;
[tex]56\sqrt[]{2}[/tex]Explanation:
Given the square in the attached image.
The length of the diagonal is;
[tex]d=28[/tex]Let l represent the length of the sides;
[tex]\begin{gathered} l^2+l^2=28^2 \\ 2l^2=784 \\ l^2=\frac{784}{2} \\ l^2=392 \\ l=\sqrt[]{392} \\ l=14\sqrt[]{2} \end{gathered}[/tex]The perimeter of a square can be calculated as;
[tex]\begin{gathered} P=4l \\ P=4(14\sqrt[]{2}) \\ P=56\sqrt[]{2} \end{gathered}[/tex]Therefore, the exact perimeter of the square is;
[tex]56\sqrt[]{2}[/tex]Use quadratic regression to find the equation of a quadratic function that fits the given points. x 0 1 2 3. Y. 49 50.4 39.5. 21
The regression Quadratic equation y = 49 + 7.55x - 6.15x².
What is Regression Equation?The technique of finding the equation of a parabola that most closely matches a collection of data is known as quadratic regression. The graph points that make up the parabola-shaped shape of this set of data are given. The parabola's equation is written as y = ax² + bx + c, where a never equals zero.
For data presented as ordered pairs, you can calculate the model's degree by identifying differences between dependent values. The model will be linear if the initial difference has the same value. The model will be quadratic if the second difference has the same value as the first.
As, we know the Quadratic model
Quadratic model, y = a + bx + cx²
Now, value of
a = 49
b = 7.55
c = -6.15
Then, the Regression Quadratic model is
y = 49 + 7.55x - 6.15x²
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Use the figures to estimate the area under the curve for the given function using four rectangles.
To calculate the area for the upper (left) graph, we can use x = 1, 2, 3 and 4 to find the upper limit of each rectangle:
[tex]\begin{gathered} f(1)=\frac{3}{1}+3=6\\ \\ f(2)=\frac{3}{2}+3=4.5\\ \\ f(3)=\frac{3}{3}+3=4\\ \\ f(4)=\frac{3}{4}+3=3.75 \end{gathered}[/tex]Since the x-interval of each rectangle is 1 unit, the area of each rectangle is given by its y-value, so we have:
[tex]\begin{gathered} A=f(1)+f(2)+f(3)+f(4)\\ \\ A=6+4.5+4+3.75=18.25 \end{gathered}[/tex]Now, for the bottom (right) graph, the limits of the rectangles are x = 2, 3, 4 and 5.
So, let's find the value of f(5):
[tex]f(5)=\frac{3}{5}+3=3.6[/tex]So the area is given by:
[tex]\begin{gathered} A=f(2)+f(3)+f(4)+f(5)\\ \\ A=4.5+4+3.75+3.6=15.85 \end{gathered}[/tex]Is the slope the same or different?Is the Y-intercept same or different?Is there infinitely many solutions or not?
Answer:
Explanation:
Here, we want to answer the questions given
a) To answer this, we have to write the equations in the slope-intercept form:
The slope-intercept form is:
[tex]y\text{ = mx + b}[/tex]m is the slope while b is the y-intercept
The equations would be:
[tex]\begin{gathered} y\text{ = 7x-2} \\ y\text{ = 7x-2} \end{gathered}[/tex]We can see that the equations are same
Since the equations are same, the slope is same which is 7
b) The y-intercept value is same too
c) Since the equations are same, there are infinitely many solutions for the system of equations
5. There are 9.75 ounces of Cinnamon Toast Crunch in a bowl. Additional cereal ispoured into the bowl at a rate of 1.5 ounces per second. How many ounces are inthe bowl after 3 seconds?
Question:
There are 9.75 ounces of Cinnamon Toast Crunch in a bowl. Additional cereal is poured into the bowl at a rate of 1.5 ounces per second. How many ounces are in the bowl after 3 seconds?
Solution:
If additional cereal is poured into the bowl at a rate of 1.5 ounces per second, then in 3 seconds the additional cereal into the bowl is 1.5x 3 = 4.5 ounces. Thus after 3 seconds, the bowl has the original amount that it already had and the new aggregate:
9.75 ounces + 4.5 ounces = 14.25
then, the correct answer is:
14.25
Leo is constructing a tangent line from point Q to circle P. What is his next step? Mark the point of intersection of circle P and segment PQ. Construct arcs from point P that are greater than half the length of segment PQ. Construct a circle from point Q with the radius PQ. Plot a new point R and create and line perpendicular to segment PQ from point R
The next step to constructing a tangent line from Q to circle P is to construct the perpendicular bisector of the segment PQ.
For this, Leo can construct arcs from point P and from point Q that are greater than half the length of segment PQ.
AnswerThe next step is to construct arcs from point P that are greater than half the length of segment PQ.
what are three requirements for fully defining a reference point?
1 - reference point should consist of abstract coordinates.
2- it should be stationary
3- it should be related to all the variables in the frame.
[tex]f(x) = (x - 2) ^{2}(x + 3)(x + 1)^{2} [/tex]the multiplicity of the root x=2 is...?
The solution of the factor with power 2 in the function f(x) can be found as:
(x-2)=0
x=2.
So, the root is x=2.
The multiplicity is the power of the factor (x-2) with its root given as x=2.
So, the multiplicity of the root x=2 is 2.
Describe the transformation from the graph of f to the graph of h. Write an equation that represents h in terms of x. Look at image for example. Let’s do problem number 11
SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given functions.
[tex]\begin{gathered} f(x)=-(x+5)^2-6 \\ h(x)=\frac{1}{3}f(x) \end{gathered}[/tex]STEP 2: Explain the transformation that occurs
What are Vertical Stretches and Shrinks?
While translations move the x and y intercepts of a base graph, stretches and shrinks effectively pull the base graph outward or compress the base graph inward, changing the overall dimensions of the base graph without altering its shape. When a graph is stretched or shrunk vertically, the x -intercepts act as anchors and do not change under the transformation.
This can be explained further as:
For the base function f (x) and a constant k > 0, the function given by:
[tex]\begin{gathered} h(x)=k\cdot f(x) \\ A\text{ vertical shrinking of f\lparen x\rparen by k factor where }0Calculate the equation that represents h in terms of x[tex]\begin{gathered} f(x)=-(x+5)^2-6 \\ h(x)=\frac{1}{3}\cdot f(x)=\frac{1}{3}\cdot-(x+5)^2-6 \end{gathered}[/tex]Hence, the transformation from the graph is a vertical shrinking by 1/3 factor and the equation that represents h in terms of x is given as:
[tex]\frac{1}{3}\times(-(x+5)^2-6)[/tex]The graph below and to the left shows the time of sunsets occurring every other day during September in a certain town. The graph at the lower right shows the time of sunsets on either the 21st or 22nd day of each month for an entire year in the same town. The vertical axis is scaled to reflect hours after midnight. Round to 4 decimal places. a) Find a linear model for the data in the graph at the left. Include units to your variables. b) Find a cosine model for the data in the graph to the right. Include units to your variables,
A) Given the points (1,18.35) and (29,17.5), we can find the linear model with the following formulas:
[tex]\begin{gathered} \text{slope:} \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{71.5-18.35}{29-1}=\frac{-0.85}{28}=-0.03 \\ \text{equation of the line:} \\ y-y_1=m(x-x_1) \\ \Rightarrow y-18.35=-0.03(x-1)=-0.03x+0.03 \\ \Rightarrow y=-0.03x+0.03+18.35=-0.03x+18.38 \\ y=-0.03x+18.38 \end{gathered}[/tex]therefore, the linear model is y = -0.03x+18.38
B)We have the general cosine model:
[tex]y(t)=A+B\cos (\omega(t-\phi))[/tex]Where A is the vertical shift, B is the amplitude, w is the frequency and phi is the phase shift.
First, we can find the vertical shift with the following formula:
[tex]A=\frac{y_{\max }+y_{\min }}{2}[/tex]in this case, we have that the maximum value for y is 19.47 and the minimum value for y is16.18, then:
[tex]A=\frac{19.47+16.18}{2}=17.825[/tex]next, we can find the amplitud with the following formula:
[tex]B=y_{\max }-A[/tex]We have then:
[tex]B=19.47-17.825=1.645[/tex]Now, notice that the graph will repeat every 356 values for t, then, for the frequency we have the following expression:
[tex]\omega=\frac{2\pi}{356}=\frac{\pi}{178}[/tex]To find the phase shift, notice that for the point (172,19.47), we have the following:
[tex]\begin{gathered} y(172)=19.47 \\ \Rightarrow17.825+1.645\cos (\frac{\pi}{178}(172-\phi))=19.47 \\ \Rightarrow1.645\cos (\frac{\pi}{178}(172-\phi))=1.645 \\ \Rightarrow\cos (\frac{\pi}{178}(172-\phi))=1 \end{gathered}[/tex]notice that if the cosine equals 1, then its argument must equal to 0, then, we have:
[tex]\begin{gathered} \frac{\pi}{178}(172-\phi)=0 \\ \Rightarrow172-\phi=0 \\ \Rightarrow\phi=172 \end{gathered}[/tex]we have that the phase shift is phi = 172, then, the final cosine model is:
[tex]y(x)=17.825+1.465\cos (\frac{\pi}{178}(x-172))[/tex]the bears have won 7 and tied 2 of their last 13 games. the not forfeited any games . which ratio correctly campares their to losses
Explanation:
Number of games won = 7
Number of games drawn = 2
Total number of games = 13
Number of games lost = Total number of games - (Number of games won + Number of games drawn)
Number of games lost = 13 - (7 + 2) = 13 - 9
Number of games lost = 7
The ratio of
the volume v of a fixed amount of a gas variety directly as the temperature T and inversely as the pressure P. suppose that V =42cm3 when T=84 kelvin and P=8kg/cm2 find the temperature when V =74cm3 and P=10 kg/cm2
ok
I'll use the law of gases to solve this problem
V1 = 42 cm^3
T1 = 84 °K
P = 8 Kg/cm^2
T2 = x
V2 = 74 cm^3
P2 = 10 kg/cm^2
Equation
P1V1/T1 = P2V2/T2
Solve for T2
T2 = P2V2T1 / P1V1
Substitution
T2 = (10*74*84) / (8*42)
Simplification
T2 = 62160 / 336
Result
T2 = 185°K
6. Write a quadratic function whose graph has a vertex of (-4,-2) and passes through the point (-3,1).
(h,k) are the coordinates of the vertex.
Use the given point (x,y) and the vertex (h,k) in the equation above to find the value of a:
Point: (-3 ,1) x = -3 y=1
Vertex: (-4 , -2) h= -4 k=-2
[tex]\begin{gathered} 1=a(-3-(-4))^2+(-2) \\ 1=a(-3+4)^2-2 \\ 1=a(1)^2_{}-2 \\ 1=a-2 \\ 1+2=a \\ 3=a \end{gathered}[/tex]Use the vertex and a to write the equation:
[tex]\begin{gathered} y=3(x-(-4))^2+(-2) \\ \\ \\ y=3(x+4)^2-2 \end{gathered}[/tex]