Question

Find the x and y intercepts of 5x^2 + 20x

Answer 1

The equation is:
\(y=5x^2+20x\)

Remember, in the \(y-intercept\), the \(x\) value is 0
In the \(x-intercept\), the \(y\) value is 0

To find the \(y-intercept\), substitute 0 for \(x\)
\(y=5x^2+20x\)
\(y=5*0^2+20*0\)
\(y=5*0+0\)
\(y=0\)

To find the \(x-intercept\), substitute 0 for \(y\)
\(y=5x^2+20x\)
\(0=5*x^2+20x\)

Then, you can either factorise, or use the quadratic equation:
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Sub in the values:
\[x=\frac{-20\pm\sqrt{20^2-4*5*0}}{2*5}\]

That will give you the two \(x-intercepts\)

Answer 2

Do you understand, @Dholmes44?