Given: f(x) = x^2 + 6x + 5 Find the x value of the vertex (turning point). X = -

Question

Given: f(x) = x^2 + 6x + 5 Find the x value of the vertex (turning point). X = -3 None of the above x = -2 x = 3 Using the same function as in #1, find the y value of the vertex. None of the above y = -4 y = 32 y = 4 Is this vertex you found above a minimum point or a maximum point? Maximum point Minimum point What is the y-intercept of the function in #1? (5, 0) None of the above (0, 5) (0, 6) What are the zeros of the parabola in #1? (that is, at what point does the graph cross the x-axis?) {(1, 0), (5, 0)} None of the above {(-1, 0), (-5, 0)} {(0, -1), (0, -5)}

Explanation / Answer

1) x value of the vertex = -b / 2a

= -6 / 2 = -3

2) y value of the vertex = (-3)2 + 6(-3) + 5

= 9 - 18 + 5 = -4

3) The vertex we have found is the minimum point

4) For y-intercept set x = 0

So, y = 02 +6*0 + 5 = 5

So the y-intercept is (0,5)

5) To find the points which cross the x-axis set y = 0

0 = x2 + 6x +5

Or, 0 = (x + 1)(x + 5)

x = -1 or x = -5

So the x-intercept points are (-1,0), ( -5,0)

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