Chapter 5.7, Problem 38PPS

To determine

To write: the standard form of quadratic equation in the vertex form as mentioned in the problem.

Answer to Problem 38PPS

The vertex form of quadratic equation is $y=-3{\left(x-5\right)}^2+4$.

Explanation of Solution

Given information:

The points on the graph is mentioned below.

  

Formula used:

The vertex form of the quadratic equation is $f\left(x\right)=a{\left(x-h\right)}^2+k$ where $\left(h,k\right)$ is the vertex.

Calculation:

Consider the points mentioned in the graph below.

  

Recall that the vertex form of the quadratic equation is $f\left(x\right)=a{\left(x-h\right)}^2+k\mathrm{......}\left(i\right)$ where $\left(h,k\right)$ is the vertex.

As the graph mentioned above we observe that vertex is $\left(5,4\right)=\left(h,k\right)$.

Therefore the equation $\left(i\right)$ becomes $y=a{\left(x-5\right)}^2+\mathrm{4.....}\left(ii\right)$.

Now since the point $\left(6,1\right)$ lies on the graph.

Therefore $x=6\&y=1$ will be point which lie on $\left(ii\right)$.

Therefore equation $\left(ii\right)$ becomes $1=a{\left(6-5\right)}^2+4$.

Therefore $1=a{\left(6-5\right)}^2+4\Rightarrow 1=a*1+4\Rightarrow 1=a+4\Rightarrow a=-3$.

Therefore the vertex form of the quadratic equation is $y=-3{\left(x-5\right)}^2+4$.