Chapter 4.2, Problem 33PPSE

Solution

To find: The equation of a parabola whose vertex and y -intercept is given.

The required parabola is $y=-\frac{4}{9}{\left(x-3\right)}^2+6$ .

Given data: The vertex of the parabola is $\left(\text{3,6}\right)$ and y -intercept 2.

Method/Formula used:

The standard equation of a parabola with vertex $\left(h\text{,}k\right)$ is given by

  $y=a{\left(x-h\right)}^2+k$ … (1)

Calculation:

The vertex of the parabola is $\left(\text{3,6}\right)$ . Therefore, from result (1), the equation of required parabola is,

  $y=a{\left(x-3\right)}^2+6$ … (2)

The y -intercept of the parabola is 2, therefore the parabola (2), passes through the point $\left(\text{0,2}\right)$ .So point $\left(\text{0,2}\right)$ should satisfy the equation (2).

Substitute $x\text{=0,}y\text{=2}$

  $\begin{array}{c}2=a{\left(0-3\right)}^2+6\\ 2-6=9a\\ a=-\frac{4}{9}\end{array}$

Substitute $-\frac{4}{9}$ for ‘ a’ in equation (2),

  $y=-\frac{4}{9}{\left(x-3\right)}^2+6$

Thus, the required parabola is $y=-\frac{4}{9}{\left(x-3\right)}^2+6$ .

The graph of required parabola is shown in Fig. (1).

  

Conclusion:

The required parabola is $y=-\frac{4}{9}{\left(x-3\right)}^2+6$ .