Chapter 4.7, Problem 4CYU

To determine

To write: the equation of the graph as mentioned.

Answer to Problem 4CYU

The equation of the graph mentioned is $y=-{\left(x+3\right)}^2+6$ .

Explanation of Solution

Given information:

The graph mentioned in the problem is as shown below.

  

Formula:

The vertex of the quadratic equation always perpendicular to x axis and the vertex is the maximum point on the graph.

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

Calculation:

Consider the graph as mentioned below.

  

Recall that vertex of the quadratic equation always perpendicular to x axis and the vertex is the maximum point on the graph.

Therefore the vertex will be $\left(-3,6\right)$ .

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

Therefore $y=a{\left(x+3\right)}^2+6$ is vertex form of the graph.

Now the point mentioned on the graph is $\left(-5,2\right)$ .

Therefore at $\left(-5,2\right)$ the value of $a$ will be $2=a{\left(-5+3\right)}^2+6\Rightarrow 2=4a+6\Rightarrow 4a=-4\Rightarrow a=-1$ .

Therefore the equation of the mentioned graph above is $y=-1{\left(x+3\right)}^2+6$ .