Question

(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 1

Answer:

(i). $y = 2\, x^2 - 4\, x + 1 = 2\, (x - 1)^2 - 1$. Point $A$ is at $(1, \, -1)$.

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

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

Step-by-step explanation:

Coordinates of the Extrema

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

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

• For the $x^2$ coefficients: $a = 2$.
• For the $x$ coefficients: $2\, a\, b = - 4$. Since $a = 2$, solving for $b$ gives $b = -1$.
• For the constant terms: $a \, b^2 + c = 1$. Since $a = 2$ and $b = -1$, solving for $c$ gives $c =-1$.

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

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

Coordinates of the Two Intersections

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

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

Take the sum of these two equations to eliminate the variable $n$:

$n + (m - n + 4) = 2\, m^2 - 4\, m + 1$.

Simplify and solve for $m$:

$2\, m^2 - 5\, m -3 = 0$.

$(2\, m + 1)\, (m - 3) = 0$.

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

• $\displaystyle m = -\frac{1}{2}$ corresponds to $n = \displaystyle \frac{7}{2}$.
• $m = 3$ corresponds to $n = 7$.

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

Line Joining Point Q and the Midpoint of Segment AP

The coordinates of point $A$ and point $P$ each have two components.

• For point $A$, the $x$-component is $1$ while the $y$-component is $(-1)$.
• For point $P$, the $x$-component is $3$ while the $y$-component is $7$.

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

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

Hence, the midpoint of segment $AP$ would be at $(2,\, 3)$.

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

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

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

$\displaystyle \left( y - \frac{7}{2}\right) = \frac{1}{5}\, \left(x - \frac{1}{2} \right)$.

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

$\displaystyle y= - \frac{1}{5}\, x + \frac{17}{5}$.

$x + 5\, y - 17 = 0$.