Chapter 9.2, Problem 8BGP

To determine

To formulate a quadratic equation from the given vertex and a point.

Answer to Problem 8BGP

The quadratic equation formed is $f(x)=\frac{5}{16}{(x)}^2+3$

Explanation of Solution

Given information :

The vertex and point on parabola provided are V $(0,3)$ and P $(4,8)$ respectively.

Formula used :

The graph of a quadratic equation ( $y=a{x}^2+bx+c$ ) is a parabola. The parabola is symmetric in nature and the point of symmetry is known as the vertex. This vertex lies on the line of symmetry and is symmetric about this line.

Here, vertex is $(h,k)$ . A quadratic equation can be defined as :-

  $f(x)=a{(x-h)}^2+k$

Calculation :

Substituting the values $h=0$ and $k=3$ .

  $f(x)=a{(x-0)}^2+3$

Substituting $x=4$ and $f(x)=8$ as point P falls on the parabola.

  $8=a{(4-0)}^2+3$

Simplifying.

  $5=16a$

Thus, $a=\frac{5}{16}$

Hence, the quadratic equation formed after substituting value of a is.

  $f(x)=\frac{5}{16}{(x)}^2+3$