Question

Consider the given function.

$y=-x^2-14x-16$

a. Determine whether the function has a maximum or minimum value.

b. State the maximum or minimum value.

c. What are the domain and range of the function?

Short answer

a. The function $y=-x^2-14x-16$ has a maximum valueat $x=-7$.

b. The maximum value of the function $y=-x^2-14x-16$ is33.
c. The domain is $\left(-\infty ,\infty \right)$and the range is $\left(-\infty ,\left.33\right]\right.$.

Step-by-step solution

Part a. Step 1. Define the standard form of the quadratic function.

A quadratic function, which is written in the form, $y=a{x}^2+bx+c$, where,$a\ne 0$ is called the standard form of the quadratic function.

Part a. Step 2. Define the maximum or minimum point of the function y=ax2+bx+c.

The graph of the function$y=a{x}^2+bx+c$

Opens upward and has a minimum value at $x=-\frac{b}{2a}$, when $a>0$.

Opens downward and has a maximum value at $x=-\frac{b}{2a}$, when $a<0$.

Part a. Step 3. Determine whether the function y=−x2−14x−16 has a maximum or minimum value.

Compare the quadratic function$y=-x^2-14x-16$ with the standard quadratic function $y=a{x}^2+bx+c$.

$a=-1,b=-14,c=-16$

Substitute$a=-1$ and$b=-14$ in $x=-\frac{b}{2a}$.

$$\begin{gathered} x=-\left(\frac{-14}{2\left(-1\right)}\right) \\ x=-\left(\frac{-14}{-2}\right) \\ x=-\left(7\right) \\ x=-7 \end{gathered}$$

Since, $a<0$.

Hence, the graph of the functionrole="math" localid="1647869649111" $y=-x^2-14x-16$opens downward and has a maximum value at$x=-7$.

Therefore the function $y=-x^2-14x-16$has a maximum value at $x=-7$.

Part b. Step 1. Define the standard form of the quadratic function.

A quadratic function, which is written in the form, $y=a{x}^2+bx+c$, where,$a\ne 0$ is called the standard form of the quadratic function.

Part b. Step 2. Define the maximum or minimum point of the function y=ax2+bx+c.

The graph of the function$y=a{x}^2+bx+c,$

Opens upward and has a minimum value at $x=-\frac{b}{2a}$, when $a>0$.

Opens downward and has a maximum value at $x=-\frac{b}{2a}$, when $a<0$.

Part b. Step 3. Calculate the maximum or minimum value of the function y=−x2−14x−16.

Compare the quadratic function$y=-x^2-14x-16$ with the standard quadratic function $y=-x^2-14x-16$.

$a=-1,b=-14,c=-16$

Substitute$a=-1$ and$b=-14$ in $x=-\frac{b}{2a}$.

$$\begin{gathered} x=-\left(\frac{-14}{2\left(-1\right)}\right) \\ x=-\left(\frac{-14}{-2}\right) \\ x=-\left(7\right) \\ x=-7 \end{gathered}$$

Since, $a<0$.

Hence, the graph of the function $y=-x^2-14x-16$opens downward and has a maximum value at $x=-7$.

Substitute$x=-7$ in $y=-x^2-14x-16$.

$$\begin{gathered} y=-{\left(-7\right)}^2-14\left(-7\right)-16 \\ y=-\left(49\right)+98-16 \\ y=-49+98-16 \\ y=-65+98 \\ y=33 \end{gathered}$$

Therefore the maximum value of the function$y=-x^2-14x-16$ is 33.

Part c. Step 1. Define the standard form of the quadratic function.

A quadratic function, which is written in the form, $y=a{x}^2+bx+c$, where,$a\ne 0$ is called the standard form of the quadratic function.

Part c. Step 2. Define the domain and range of the function.

The domain is the set of all of the possible values of the independent variable $x$.

The range is the set of all the possible values of the dependent variable $y$.

Part c. Step 3. Determine the domain and range of the function y=−x2−14x−16.

Since, the graph of the function$y=-x^2-14x-16$ is a parabola.

Since, the parabola always extends to infinity.

So, the domain is $\left(-\infty ,\infty \right)$.

Since, the maximum value of the function is 33.

So, the range is $\left(-\infty ,\left.33\right]\right.$.

Therefore, the domain is$\left(-\infty ,\infty \right)$ and the range is $\left(-\infty ,\left.33\right]\right.$.