Question

Algebra 1 semester 2

Answer 1

1) The x intercept is the value of x when y is zero

The given equation is expressed as

y = x^2 - 4x - 12

To find the x intercepts, we would make y = 0

Thus, we have

x^2 - 4x - 12 = 0

This is a quadratic equation. The general form of a quadratic equation is expressed as

ax^2 + bx + c = 0

By comparing,

a = 1, b = - 4, c = - 12

The formula for solving quadratic equations is expressed as

`\begin{gathered} x\text{ = }\frac{-\text{ b }\pm\sqrt[]{b^2-4ac}}{2a} \\ By\text{ substituting, } \\ x\text{ = }\frac{-\text{ - 4 }\pm\sqrt[]{-4^2\text{ - 4(1 }*\text{ - 12)}}}{2\text{ }*1} \\ x\text{ = }\frac{4\text{ }\pm\sqrt[]{16\text{ + 48}}}{2} \\ x\text{ = }\frac{4\text{ }\pm\text{ }\sqrt[]{64}}{2} \\ x\text{ = }\frac{4\text{ + 8}}{2}\text{ or x = }\frac{4\text{ - 8}}{2} \\ x\text{ = 6 or x = - 2} \end{gathered}`

The x intercepts are x = 6 or x = - 2