(i)Express 2x² – 4x + 1 in the form a(x+ b)² + c and hence state the coordinates of the minimum point, A, on the curve y= 2x² 4x+ 1.
The line x– y + 4 = 0 intersects the curve y= 2x² – 4x + 1 at points P and Q. It is given that the coordinates of Pare (3,7).

(ii)Find the coordinates of
(iii) Find the equation of the line joining Q to the mid-point of AP.

I already get the (i) but hav no idea with no.(ii) and (iii)

Can anyone help me? Thank you​

Answer:

(i). [tex]y = 2\, x^2 - 4\, x + 1 = 2\, (x - 1)^2 - 1[/tex]. Point [tex]A[/tex] is at [tex](1, \, -1)[/tex].

(ii). Point [tex]Q[/tex] is at [tex]\displaystyle \left(-\frac{1}{2},\, \frac{7}{2}\right)[/tex].

(iii). [tex]\displaystyle y= - \frac{1}{5}\, x + \frac{17}{5}[/tex] (slope-intercept form) or equivalently [tex]x + 5\, y - 17 = 0[/tex] (standard form.)

Step-by-step explanation:

Note, that when [tex]a(x + b)^2 + c[/tex] is expanded, the expression would become [tex]a\, x^2 + 2\, a\, b\, x + a\, b^2 + c[/tex].

Compare this expression to the original [tex]2\, x^2 - 4\, x + 1[/tex]. In particular, try to match the coefficients of the [tex]x^2[/tex] terms and the [tex]x[/tex] terms, as well as the constant terms.

Hence, the original expression for the parabola is equivalent to [tex]y = 2\, (x - 1)^2 - 1[/tex].

For a parabola in the vertex form [tex]y = a\, (x + b)^2 + c[/tex], the vertex (which, depending on [tex]a[/tex], can either be a minimum or a maximum,) would be [tex](-b,\, c)[/tex]. For this parabola, that point would be [tex](1,\, -1)[/tex].

Assume [tex](m,\, n)[/tex] is an intersection of the graphs of the two functions [tex]y = 2\, x^2- 4\, x + 1[/tex] and [tex]x -y + 4 = 0[/tex]. Setting [tex]x[/tex] to [tex]m[/tex], and [tex]y[/tex] to [tex]n[/tex] should make sure that both equations still hold. That is:

[tex]\displaystyle \left\lbrace \begin{aligned}& n = 2\, m^2 - 4\, m + 1 \\ & m - n + 4 = 0\end{aligned}\right.[/tex].

Take the sum of these two equations to eliminate the variable [tex]n[/tex]:

[tex]n + (m - n + 4) = 2\, m^2 - 4\, m + 1[/tex].

Simplify and solve for [tex]m[/tex]:

[tex]2\, m^2 - 5\, m -3 = 0[/tex].

[tex](2\, m + 1)\, (m - 3) = 0[/tex].

There are two possible solutions: [tex]m = -1/2[/tex] and [tex]m = 3[/tex]. For each possible [tex]m[/tex], substitute back to either of the two equations to find the value of [tex]n[/tex].

Hence, the two intersections are at [tex]\displaystyle \left(-\frac{1}{2},\, \frac{7}{2}\right)[/tex] and [tex](3,\, 7)[/tex], respectively.

The coordinates of point [tex]A[/tex] and point [tex]P[/tex] each have two components.

Let [tex]M[/tex] denote the midpoint of segment [tex]AP[/tex]. The [tex]x[/tex]-component of point [tex]M[/tex] would be [tex](1 + 3) / 2 = 2[/tex], the average of the [tex]x[/tex]-components of point [tex]A[/tex] and point [tex]P[/tex].

Similarly, the [tex]y[/tex]-component of point [tex]M[/tex] would be [tex]((-1) + 7) / 2 = 3[/tex], the average of the [tex]y\![/tex]-components of point [tex]A[/tex] and point [tex]P[/tex].

Hence, the midpoint of segment [tex]AP[/tex] would be at [tex](2,\, 3)[/tex].

The slope of the line joining [tex]\displaystyle \left(-\frac{1}{2},\, \frac{7}{2}\right)[/tex] (the coordinates of point [tex]Q[/tex]) and [tex](2,\, 3)[/tex] (the midpoint of segment [tex]AP[/tex]) would be:

[tex]\displaystyle \frac{\text{Change in $y$}}{\text{Change in $x$}} = \frac{3 - (7/2)}{2 - (-1/2)} = \frac{1}{5}[/tex].

Point [tex](2,\, 3)[/tex] (the midpoint of segment [tex]AP[/tex]) is a point on that line. The point-slope form of this line would be:

[tex]\displaystyle \left( y - \frac{7}{2}\right) = \frac{1}{5}\, \left(x - \frac{1}{2} \right)[/tex].

Rearrange to obtain the slope-intercept form, as well as the standard form of this line:

[tex]\displaystyle y= - \frac{1}{5}\, x + \frac{17}{5}[/tex].

[tex]x + 5\, y - 17 = 0[/tex].

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