Given the function:
[tex]f(x)=x^2+4x-12[/tex]
Let's graph the function.
Let's find the following:
• (a). x-intercepts:
The x-intercepts are the points the function crosses the x-axis.
To find the x-intercepts substitute 0 for f(x) and solve for x.
[tex]\begin{gathered} 0=x^2+4x-12 \\ \\ x^2+4x-12=0 \end{gathered}[/tex]
Factor the left side using AC method.
Find a pair of numbers whose sum is 4 and product is -12.
We have:
6 and -2
Hence, we have
[tex]\begin{gathered} (x+6)(x-2)=0 \\ \\ \end{gathered}[/tex]
Equate the individual factors to zero and solve for x.
[tex]\begin{gathered} x+6=0 \\ Subtract\text{ 6 frm both sides:} \\ x+6-6=0-6 \\ x=-6 \\ \\ \\ x-2=0 \\ Add\text{ 2 to both sides:} \\ x-2+2=0+2 \\ x=2 \end{gathered}[/tex]
Therefore, the x-intercepts are:
x = -6 and 2
In point form, the x-intercepts are:
(x, y) ==> (-6, 0) and (2, 0)
• (b). The y-intercept.
The y-intercept is the point the function crosses the y-axis.
Substitute 0 for x and solve f(0) to find the y-intercept:
[tex]\begin{gathered} f(0)=0^2+4(0)-12 \\ \\ f(0)=-12 \end{gathered}[/tex]
Therefore, the y-intercept is:
y = -12
In point form, the y-intercept is:
(x, y) ==> (0, -12)
• (c). What is the maximum or minimum value?
Since the leading coefficient is positive the graph will have a minimum value.
To find the point where it is minimum, apply the formula:
[tex]x=-\frac{b}{2a}[/tex]
Where:
b = 4
a = 1
Thus, we have:
[tex]\begin{gathered} x=-\frac{4}{2(1)} \\ \\ x=-\frac{4}{2} \\ \\ x=-2 \end{gathered}[/tex]
To find the minimum values, substitute -2 for x and solve for f(-2):
[tex]\begin{gathered} f(-2)=(-2)^2+4(-2)-12 \\ \\ f(-2)=4-8-12 \\ \\ f(-2)=-16 \end{gathered}[/tex]
Therefore, the minimum value is at:
y = -16
Using the point form, we have the minimum point:
(x, y) ==> (-2, -16).
• (d). Use the points to plot the graph.
We have the points:
(x, y) ==> (-6, 0), (2, 0), (0, -12), (-2, -16)
Plotting the graph using the points, we have:
