Chapter SH, Problem 4.20EP

To determine

Write an equation of the given parabola in vertex form.

Answer to Problem 4.20EP

The equation of the parabola is $y={\left(x-3\right)}^2+2$ .

Explanation of Solution

Given:

The given parabola is

  

Calculation:

Equation in vertex form of the parabola is

  $y=a{\left(x-h\right)}^2+k$ , where $\left(h,k\right)$ is the vertex of the parabola, $x=h$ is the axis of symmetry, and ‘a’ determines the shape of the parabola and the direction in which it opens.

If $a>0$ , the graph opens up.

If $a<0$ , the graph opens down.

Vertex $\left(h,k\right)=\left(3,2\right)$

Use vertex form of the equation $y=a{\left(x-h\right)}^2+k$ .

  $y=a{\left(x-3\right)}^2+2$

Point on the graph $=\left(1,6\right)$

Substitute the values of $x=1$ and $y=6$ in the equation and find the value of ‘a’.

  $\begin{array}{l}\Rightarrow 6=a{\left(1-3\right)}^2+2\\ \Rightarrow 6=4a+2\\ \Rightarrow 6-2=4a+2-2\\ \Rightarrow \frac{4}{4}=\frac{4}{4}a\\ \Rightarrow 1=a\\ \Rightarrow a=1\end{array}$

Hence the equation of the parabola is $y={\left(x-3\right)}^2+2$ .