Question

use the intercepts and extreme value to find an equation for the quadratic function.
a f(x) = 2x² - 3x - 6
b f(x) = 3x² - 4x - 6
c f(x) = 2x² - 4x - 6
d f(x) = 2x² - 4x + 6

Explanation:

Step1: Check y-intercept

The y-intercept occurs at \( x = 0 \). From the graph, when \( x = 0 \), \( y=-6 \). Let's check each option:

• Option A: \( f(0)=2(0)^2 - 3(0)-6=-6 \) (good)
• Option B: \( f(0)=3(0)^2 - 4(0)-6=-6 \) (good)
• Option C: \( f(0)=2(0)^2 - 4(0)-6=-6 \) (good)
• Option D: \( f(0)=2(0)^2 - 4(0)+6 = 6\) (bad, eliminate D)

Step2: Check x-intercepts

From the graph, x-intercepts are at \( x=-1 \) and \( x = 3 \). Let's check these in remaining options (A, B, C).

• For \( x=-1 \):
• Option A: \( f(-1)=2(-1)^2-3(-1)-6=2 + 3-6=-1

eq0 \) (bad)

• Option B: \( f(-1)=3(-1)^2-4(-1)-6=3 + 4-6 = 1

eq0 \) (bad)

• Option C: \( f(-1)=2(-1)^2-4(-1)-6=2 + 4-6 = 0\) (good)
• For \( x = 3 \):
• Option C: \( f(3)=2(3)^2-4(3)-6=18 - 12-6 = 0\) (good)

Step3: Verify extreme value (optional, but confirms)

The vertex form of a quadratic is \( f(x)=a(x - h)^2+k \). From x-intercepts \( x=-1 \) and \( x = 3 \), the axis of symmetry is \( x=\frac{-1 + 3}{2}=1 \). For option C, \( f(x)=2x^2-4x - 6 \), rewrite in vertex form: \( f(x)=2(x^2-2x)-6=2(x^2-2x + 1-1)-6=2(x - 1)^2-2 - 6=2(x - 1)^2-8 \). The vertex is at \( (1,-8) \), which matches the graph's minimum point.

Answer:

C. \( f(x) = 2x^2 - 4x - 6 \)