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Likurg_2 [28]
1 year ago
9

(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​

Mathematics
1 answer:
earnstyle [38]1 year ago
7 0

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:

<h3>Coordinates of the Extrema</h3>

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).

<h3>Coordinates of the Two Intersections</h3>

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.

<h3>Line Joining Point Q and the Midpoint of Segment AP</h3>

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.

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