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

(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(-(1)/(2),\, (7)/(2)\right)`.

(iii).
`\displaystyle y= - (1)/(5)\, x + (17)/(5)` (slope-intercept form) or equivalently
`x + 5\, y - 17 = 0` (standard form.)

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 = -(1)/(2)` corresponds to
  `n = \displaystyle (7)/(2)`.
• 
  `m = 3` corresponds to
  `n = 7`.

Hence, the two intersections are at
`\displaystyle \left(-(1)/(2),\, (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(-(1)/(2),\, (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 $x$}} = (3 - (7/2))/(2 - (-1/2)) = (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 - (7)/(2)\right) = (1)/(5)\, \left(x - (1)/(2) \right)`.

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

`\displaystyle y= - (1)/(5)\, x + (17)/(5)`.

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