Question

select the inequality which represents the graph shown below.
answer
\\(\circ\\) \\(y - 12 \geq x^2 + 8x\\)
\\(\circ\\) \\(y - 12 \leq x^2 - 8x\\)
\\(\circ\\) \\(y - 12 \leq x^2 + 8x\\)
\\(\circ\\) \\(y - 12 \geq x^2 - 8x\\)

Explanation:

Step1: Find roots of the parabola

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

Step2: Find the leading coefficient $a$

The y-intercept is $(0,10)$. Substitute $x=0,y=10$:
$10=a(0-2)(0-6) \implies 10=12a \implies a=\frac{10}{12}=\frac{5}{6}$? No, expand standard form:
$y=(x-2)(x-6)=x^2-8x+12$. Check y-intercept: when $x=0$, $y=12$, which matches the graph (the parabola meets y-axis at 10? No, correction: the graph's y-intercept is 10? No, recheck: the vertex is at $x=\frac{2+6}{2}=4$, substitute $x=4$ into $x^2-8x+12$: $16-32+12=-4$, which matches the vertex $(4,-4)$ on the graph. The y-intercept of $x^2-8x+12$ is 12, which matches the top of the parabola at y=10? No, correction: the graph's y-intercept is 10? No, the graph shows the parabola starting at y=10, but the vertex form $y=(x-4)^2-4 = x^2-8x+12$, which has y-intercept 12. The shaded area is outside the parabola, so $y \geq x^2-8x+12$. Rearrange: $y-12 \geq x^2-8x$.

Step3: Verify inequality direction

The shaded region is outside (above) the parabola, so $y \geq x^2-8x+12$, which rearranges to $y-12 \geq x^2-8x$.

Answer:

$\boldsymbol{y - 12 \geq x^2 - 8x}$ (the bottom-right option)