Q44. - Algebra 1

Question

Consider the given function.

$y=x^2+4x-5$

 

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 $y=x^2+4x-5$ has a minimum value at $x=-2$.

b. The minimum value of the function $y=x^2+4x-5$ is $-9$.

c. The domain is $\left(-\infty ,\infty \right)$ and the range is $\left[-9,\left.\infty \right)\right.$.

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 = 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 $a<0$.

Part a. Step 3. Determine whether the function y = x 2 + 4 x &#8722; 5 has a maximum or minimum value.

Compare the quadratic function $y=x^2+4x-5$ with the standard quadratic function $y=a{x}^2+bx+c$.

$a=1,b=4,c=-5$

Substitute $a=1$ and $b=4$ in $x=-\frac{b}{2a}$.

$$\begin{gathered} x=-\left(\frac{4}{2\left(1\right)}\right) \\ x=-\left(\frac{4}{2}\right) \\ x=-\left(2\right) \\ x=-2 \end{gathered}$$

Since, .   

Hence, the graph of the function $y=x^2+4x-5$ opens upward and has a minimum value at $x=-2$.

 

Therefore the function $y=x^2+4x-5$ has a minimum value at $x=-2$.

 

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 = 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 $a<0$.

Part b. Step 3. Calculate the maximum or minimum value of the function y = x 2 + 4 x &#8722; 5 .

Compare the quadratic function $y=x^2+4x-5$ with the standard quadratic function $y=a{x}^2+bx+c$.

$a=1,b=4,c=-5$

Substitute $a=1$ and $b=4$ in $x=-\frac{b}{2a}$.

$$\begin{gathered} x=-\left(\frac{4}{2\left(1\right)}\right) \\ x=-\left(\frac{4}{2}\right) \\ x=-\left(2\right) \\ x=-2 \end{gathered}$$

Since, $a>0$.

Hence, the graph of the function $y=x^2+4x-5$ opens upward and has a minimum value at $x=-2$. 

Substitute $x=-2$ in $y=x^2+4x-5$.

$$\begin{gathered} y={\left(2\right)}^2+4\left(-2\right)-5 \\ y=4-8-5 \\ y=4-13 \\ y=-9 \end{gathered}$$

 

Therefore the minimum value of the function $y=x^2+4x-5$ is $-9$.

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 = x 2 + 4 x &#8722; 5 .

Since, the graph of the function $y=x^2+4x-5$ is a parabola.

Since, the parabola always extends to infinity.

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

Since, the minimum value of the function is $-9$.

So, the range is $\left[-9,\left.\infty \right)\right.$.

 

Therefore, the domain is $\left(-\infty ,\infty \right)$ and the range is $\left[-9,\left.\infty \right)\right.$.