Q14. - Algebra 1

Question

Consider the given function.

$y=-x^2+2$

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? 

Step-by-Step Solution

Answer

a. The function $y=-x^2+2$ has a maximum value.

b. The maximum value of the function $y=-x^2+2$ is 2.

c. The domain is $\left(-\infty ,\infty \right)$ and the range is $(-\infty ,\left.2\right]$.

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.

Step 2. Define the maximum or minimum point of the function y = a x 2 + b x + 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 $a<0$.

Step3. calculation

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

$a=-1,b=0,c=2$

Since $a<0$.

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

Therefore, the function $y=-x^2+2$ has a maximum value.

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

$a=-1,b=0,c=2$

Substitute $a=-1$ and $b=0$ in $x=-\frac{b}{2a}$.

$$\begin{gathered} x=-\left(\frac{0}{2\left(-1\right)}\right) \\ x=0 \end{gathered}$$

Since $a<0$.

Hence, the graph of the function $y=-x^2+2$ opens downward and has a maximum value at $x=0$.

Substitute $x=0$ in $y=-x^2+2$.

$$\begin{gathered} y=-0^2+2 \\ y=0+2 \\ y=2 \end{gathered}$$

Therefore the maximum value of the function $y=-x^2+2$ is 2.

c. 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 $y$.       

Since the graph of the function $y=-x^2+2$ 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 2.

So, the range is $\left(-\infty ,\left.2\right]\right.$.

Therefore, the domain is $\left(-\infty ,\infty \right)$ and the range is $\left(-\infty ,\left.2\right]\right.$.