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H(x)=−x2−6x−16 h ( x ) = − x 2 − 6 x − 16

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Answer:

To analyze the function H(x) = -x^2 - 6x - 16, we can start by finding its vertex and intercepts:

Vertex: The x-coordinate of the vertex of the parabola represented by H(x) is given by x = -b/2a, where a and b are the coefficients of x^2 and x, respectively. In this case, a = -1 and b = -6, so x = -(-6)/2(-1) = 3. To find the corresponding y-coordinate, we substitute x = 3 into H(x) and get:

H(3) = -(3)^2 - 6(3) - 16 = -25

Therefore, the vertex of H(x) is (3, -25).

y-intercept: To find the y-intercept, we set x = 0 in H(x) and get:

H(0) = -(0)^2 - 6(0) - 16 = -16

Therefore, the y-intercept is (0, -16).

x-intercepts: To find the x-intercepts, we set H(x) = 0 and solve for x:

-x^2 - 6x - 16 = 0

Multiplying both sides by -1 and rearranging, we get:

x^2 + 6x + 16 = 0

Using the quadratic formula, we get:

x = (-6 ± √(6^2 - 4(1)(16))) / (2(1)) = (-6 ± 2√5i) / 2 = -3 ± √5i

Therefore, the x-intercepts are (-3 + √5i, 0) and (-3 - √5i, 0).

Axis of symmetry: The axis of symmetry is the vertical line passing through the vertex, which is x = 3 in this case.

Direction of opening: Since the coefficient of x^2 is negative, the parabola opens downwards.

Maximum value: Since the parabola opens downwards, the vertex represents the maximum point of the function. Therefore, the maximum value of H(x) is -25.

With this information, we can sketch the graph of H(x), which looks like a downward-facing parabola with vertex at (3, -25), x-intercepts at (-3 + √5i, 0) and (-3 - √5i, 0), and y-intercept at (0, -16).

User Udaya Sri
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