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Math  /  Algebra

QuestionSketch the graph of each inequality. 4) y≤−x2+8x−12y \leq-x^{2}+8 x-12 A) C) B) D)

Studdy Solution

STEP 1

1. The inequality y≤−x2+8x−12 y \leq -x^2 + 8x - 12 represents a region on the coordinate plane.
2. The parabola y=−x2+8x−12 y = -x^2 + 8x - 12 is the boundary of the region.
3. The inequality includes the boundary since it is ≤ \leq .

STEP 2

1. Identify the vertex and axis of symmetry of the parabola.
2. Determine the direction in which the parabola opens.
3. Sketch the parabola and shade the appropriate region.

STEP 3

To find the vertex of the parabola y=−x2+8x−12 y = -x^2 + 8x - 12 , use the vertex formula x=−b2a x = -\frac{b}{2a} .
Here, a=−1 a = -1 , b=8 b = 8 , so x=−82(−1)=4 x = -\frac{8}{2(-1)} = 4 .
Substitute x=4 x = 4 back into the equation to find y y :
y=−(4)2+8(4)−12=−16+32−12=4 y = -(4)^2 + 8(4) - 12 = -16 + 32 - 12 = 4
Thus, the vertex is (4,4) (4, 4) .

STEP 4

The coefficient of x2 x^2 is negative, so the parabola opens downwards.

STEP 5

Plot the vertex (4,4) (4, 4) on the coordinate plane.
Draw the parabola opening downwards, passing through the vertex.
Since the inequality is y≤−x2+8x−12 y \leq -x^2 + 8x - 12 , shade the region below the parabola, including the parabola itself.
The graph that matches this description is the correct sketch of the inequality. Look for the graph where the parabola opens downwards with vertex at (4,4) (4, 4) and the region below it is shaded.

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