Chapter 4.2, Problem 53PPSE

Solution

To sketch: The graph of a parabola for given axis of symmetry and point/s.

Given data: The parabola has $x=2$ as the axis of symmetry, y -intercept 1 and point $\left(3,2.5\right)$ .

Method/Formula used:

The vertex form of the parabola is given by $y=a{\left(x-h\right)}^2+k$ in which $x=h$ is the axis of symmetry and $\left(h,k\right)$ the vertex.

Calculation:

Suppose the equation of parabola is

  $y=a{\left(x-h\right)}^2+k$ . … (1)

Since axis of symmetry of the required parabola is $x=2$ , that implies $h=2$ . Therefore, the equation of required parabola is

  $y=a{\left(x-2\right)}^2+k$ . … (2)

The parabola has y -intercept equal to 1, therefore, parabola (2), passes through point $\left(0,1\right)$ ), so point $\left(0,1\right)$ should satisfy equation (2), substitute $x=0,y=1$ in equation (2),

  $\begin{array}{l}1=a{\left(0-2\right)}^2+k\\ 1=4a+k\end{array}$

OR

  $4a+k=1$ … (3)

Also, the parabola passes through the point $\left(\text{3,5/2}\right)$ , therefore, the point $\left(\text{3,5/2}\right)$ should satisfy equation (2), substitute $x=3,y=5/2$ in equation (2),

  $\begin{array}{l}2.5=a{\left(3-2\right)}^2+k\\ 2.5=a{\left(3-2\right)}^2+k\\ 2.5=a+k\end{array}$

OR

  $a+k=2.5$ … (4)

Subtraction equation (4) from equation (3),

  $\begin{array}{c}\left(4a+k\right)-\left(a+k\right)=1-2.5\\ 4a-a+k-k=-1.5\\ 3a=-1.5\\ a=-\frac{1.5}{3}\end{array}$

Therefore,

  $a=-\frac{1}{2}$

Substitute $-0.5$ for a in equation (4),

  $\begin{array}{c}-0.5+k=2.5\\ k=2.5+0.5\\ k=3\end{array}$

Substitute $-\frac{1}{2}$ for a , and 3 for k in equation (2), the required equation of parabola is $y=-\frac{1}{2}{\left(x-2\right)}^2+3$ .

The graph of the parabola is shown in Fig. 1.