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Consider h(x) = -x2 + 8x + 16. Identify its vertex and y-intercept and axis of symmetry ,range

User Gigatron
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Final answer:

The vertex of the function is (4, 48), the y-intercept is (0, 16), the axis of symmetry is x = 4, and the range is [-∞, 48].

Step-by-step explanation:

The given function is h(x) = -x^2 + 8x + 16.

To find the vertex, we first need to find the x-coordinate of the vertex, which is given by the formula x = -b / (2a) in the equation of a quadratic function y = ax^2 + bx + c. In this case, a = -1 and b = 8. Substituting these values into the formula, we get x = -8 / (2*(-1)) = 4. To find the y-coordinate of the vertex, substitute this value of x into the function, giving y = -4^2 + 8*4 + 16 = 48. Therefore, the vertex is (4, 48).

The y-intercept is the value of the function when x = 0. Substituting x = 0 into the function, we get y = -0^2 + 8*0 + 16 = 16. Hence, the y-intercept is (0, 16).

The axis of symmetry is the vertical line that passes through the vertex. In this case, the axis of symmetry is x = 4.

The range of the function is the set of all possible y-values. Since the coefficient of the x^2 term is negative, the parabola opens downwards and has a maximum value. Therefore, the range is [-∞, 48].

User James Snook
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