To find the vertex form of the function, we can complete the square. The vertex form is given by: s(x) = a(x-h)^2 + k, where (h, k) represents the coordinates of the vertex.
s(x) = x^2 - 8x + 7
s(x) = (x^2 - 8x) + 7
To complete the square, we need to add and subtract the square of half the coefficient of x:
s(x) = (x^2 - 8x + 16 - 16) + 7
s(x) = (x^2 - 8x + 16) - 16 + 7
s(x) = (x - 4)^2 - 9
Therefore, the vertex form of the function s(x) = x^2 - 8x + 7 is s(x) = (x - 4)^2 - 9.
(A) To find the x-intercepts, we set s(x) = 0 and solve for x:
(x - 4)^2 - 9 = 0
(x - 4)^2 = 9
Taking the square root of both sides:
x - 4 = ±√9
x - 4 = ±3
x = 4 ± 3
x = 7 or x = 1
Therefore, the x-intercepts are x = 7 and x = 1.
To find the y-intercept, we set x = 0:
s(0) = (0 - 4)^2 - 9
s(0) = (-4)^2 - 9
s(0) = 16 - 9
s(0) = 7
Therefore, the y-intercept is y = 7.
(B) The vertex of the parabola is given by (h, k) = (4, -9).
(C) Since the coefficient of the x^2 term is positive (1), the parabola opens upwards and the vertex represents the minimum point.
(D) The range of the function is all real numbers greater than or equal to -9, since the vertex is the minimum point. So, Range: [-9, ∞).
Find the vertex form of the function. Then find each of the following.
(A) Intercepts
(B) Vertex
(C) Maximum or minimum
(D) Range
s(x)= x^2-8x+7
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