Question

1. What are the x-intercepts ofy = 2x^2 – 3x-20?

Answer 1

The given equation is,

`y=2x^2-3x-20`

Put y=0 in the above equation and solve for x to find the x intercepts.

Putting y =0 in the above equation,

`0=2x^2-3x-20\ldots\ldots(1)_{}`

The above equation is in the form of a quadratic equation given by,

`ax^2+bx+c=0`

Comparing the equations, a=2, b=-3 and c=-20.

Now, using discriminant method solve equation (1) for x.

`\begin{gathered} x=\frac{-(-3)\pm\sqrt[]{(-3)^2-4*2*(-20)}}{2*2} \\ =\frac{3\pm\sqrt[]{9^{}+160}}{4} \\ =\frac{3\pm\sqrt[]{169}}{4} \\ =(3\pm13)/(4) \\ x=(3+13)/(4)\text{ or x=}(3-13)/(4) \\ x=(16)/(4)\text{ or x=}(-10)/(4) \\ x=4\text{ or x=}(-5)/(2) \end{gathered}`

Therefore, the x intercepts of the graph of the given equation is x=4 or x=-5/2.