Question

1. what is the equation of this quadratic?

graph of a parabola with x-intercepts at (-6, 0) and (2, 0), y-intercept at (0, -8), vertex at (-2, -8)
options:

• ( y = x^2 + 2x - 8 )
• ( y = \frac{1}{2}x^2 + 2x - 8 )
• ( y = \frac{1}{4}x^2 + 2x - 8 )
• ( y = (x - 2)(x + 6) )
• ( y = \frac{1}{2}(x - 2)(x + 6) )
• ( y = \frac{1}{4}(x - 2)(x + 6) )

Explanation:

Step1: Identify x-intercepts

The parabola crosses the x-axis at $x=2$ and $x=-6$. So the factored form is $y=a(x-2)(x+6)$, where $a$ is the leading coefficient.

Step2: Use a point to find $a$

The parabola passes through $(0,-6)$. Substitute $x=0$, $y=-6$ into the factored form:
$$-6 = a(0-2)(0+6)$$
$$-6 = a(-2)(6)$$
$$-6 = -12a$$
Solve for $a$: $a = \frac{-6}{-12} = \frac{1}{2}$

Step3: Write full equation

Substitute $a=\frac{1}{2}$ into the factored form: $y=\frac{1}{2}(x-2)(x+6)$. We can verify by expanding:
$$y=\frac{1}{2}(x^2+4x-12)=\frac{1}{2}x^2+2x-6$$
This matches the y-intercept $(0,-6)$ and the shape of the parabola.

Answer:

y = 1/2 (x - 2)(x + 6)