Question

write an equation for a parabola with x-intercepts (-2,0) and (4,0) which passes through the point (3, -10). write the equation.

Explanation:

Step1: Recall the factored form of a parabola

The factored form of a parabola with x - intercepts \(x = r_1\) and \(x = r_2\) is \(y=a(x - r_1)(x - r_2)\). Given the x - intercepts \((-2,0)\) and \((4,0)\), we have \(r_1=-2\) and \(r_2 = 4\). So the equation of the parabola in factored form is \(y=a(x+2)(x - 4)\).

Step2: Find the value of \(a\)

We know that the parabola passes through the point \((3,-10)\). Substitute \(x = 3\) and \(y=-10\) into the equation \(y=a(x + 2)(x - 4)\):
\(-10=a(3 + 2)(3-4)\)
First, calculate the values inside the parentheses: \(3 + 2=5\) and \(3-4=-1\). So the equation becomes \(-10=a\times5\times(-1)\), which simplifies to \(-10=-5a\).
To solve for \(a\), divide both sides of the equation by \(-5\): \(a=\frac{-10}{-5}=2\).

Step3: Write the final equation

Substitute \(a = 2\) back into the factored form \(y=a(x + 2)(x - 4)\). We get \(y = 2(x+2)(x - 4)\). If we expand this, we have:
First, multiply \((x + 2)(x - 4)=x^{2}-4x+2x - 8=x^{2}-2x - 8\). Then multiply by 2: \(y = 2x^{2}-4x-16\).

Answer:

\(y = 2(x + 2)(x - 4)\) (or \(y=2x^{2}-4x - 16\))