Chapter 1.7, Problem 44E

Solution

To write: The function $y=x^2+20x+90$ in vertex form and identify the vertex.

The vertex form of the function is $y={\left(x+10\right)}^2-10$ and the vertex is $\left(-10,-10\right)$ .

Given information:

The function, $y=x^2+20x+90$ .

Concept used:

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

Calculation:

Write the function in the vertex form as follows:

  $\begin{array}{c}y=x^2+20x+90\\ y=x^2+2\cdot x\cdot 10+{10}^2-{10}^2+90\\ y={\left(x+10\right)}^2-{10}^2+90\\ y={\left(x+10\right)}^2-100+90\\ y={\left(x+10\right)}^2-10\end{array}$

Now, compare the function $y={\left(x+10\right)}^2-10$ with $y=a{\left(x-h\right)}^2+k$ to get the vertex as $\left(-10,-10\right)$ .

Conclusion:

The vertex form of the function is $y={\left(x+10\right)}^2-10$ and the vertex is $\left(-10,-10\right)$ .