Answer:
Explanation:
Given the function
[tex]f\left(x\right)\:=\:5x^2-\:20[/tex]
As we know that the x-intercept(s) can be obtained by setting the value y=0
so
[tex]y=\:5x^2-\:20[/tex]
switching sides
[tex]5x^2-20=0[/tex]
Add 20 to both sides
[tex]5x^2-20+20=0+20[/tex]
[tex]5x^2=20[/tex]
Dividing both sides by 5
[tex]\frac{5x^2}{5}=\frac{20}{5}[/tex]
[tex]x^2=4[/tex]
[tex]\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=\sqrt{4},\:x=-\sqrt{4}[/tex]
[tex]x=2,\:x=-2[/tex]
so the x-intercepts will be: (2, 0), (-2, 0)
we also know that the y-intercept(s) can obtained by setting the value x=0
so
[tex]y=\:5(0)^2-\:20[/tex]
[tex]y=0-20[/tex]
[tex]y=-20[/tex]
so the y-intercept will be: (-20, 0)
From the attached figure, all the intercepts are labeled.