Question

Please Help Guys! 1) Why is f(x)=(3x+13)2+89 not the vertex form of f(x)=9x2+2x+1? A.The expression has a constant outside of the squared term. B. The expression is not the product of two binomials. C. The variable x has a coefficient. D. Some of the terms are fractions instead of integers. 2) What is the vertex of the parabola with the equation y=(x−2)2+10? A. (−2, −10) B. (2, 10) C. (−2, 10) D. (2, −10) 3) For the given function, identify the x- and y-intercepts if any, the vertex, the axis of symmetry, and the maximum or minimum value. f(x)=−x2+25 A. The x-intercepts are (−5,0) and (5,0). The y-intercept is (0,−25). The vertex is (0,−25). The axis of symmetry is x=0. The minimum value of the function is −25. B. There are no x-intercepts. The y-intercept is (0,25). The vertex is (0,25). The axis of symmetry is y=0. The maximum value of the function is 25. C. The x-intercepts are (−5,0) and (5,0). The y-intercept is (0,25). The vertex is (0,25). The axis of symmetry is x=0. The maximum value of the function is 25. D. The x-intercepts are (−25,0) and (25,0). The y-intercept is (0,5). The vertex is (0,5). The axis of symmetry is x=0. The maximum value of the function is 5. 4) A student says that the function f(x)=−x2−9 has the x-intercepts (−3,0) and (3,0). Is the student correct? If not, explain why. A. The student is correct. B. The student is not correct. The equation f(x)=0 has one real solution, so the x-intercept is (9,0). C. The student is not correct. The equation f(x)=0 does not have any real solutions, so the graph has only one x-intercept, (0,0). D. The student is not correct. The equation f(x)=0 does not have any real solutions, so the graph does not have any x-intercepts.

Answer 1

1. C. The variable x has a coefficient.
2. B. (2, 10)
3. C. The x-intercepts are (−5,0) and (5,0). The y-intercept is (0,25). The vertex is (0,25). The axis of symmetry is x=0. The maximum value of the function is 25.
4. D. The student is not correct. The equation f(x)=0 does not have any real solutions, so the graph does not have any x-intercepts.

Explanation:

1) Why is f(x)=(3x+1/3)^2+8/9 not the vertex form of f(x)=9x^2+2x+1?

A.The expression has a constant outside of the squared term.

B. The expression is not the product of two binomials.

C. The variable x has a coefficient.

D. Some of the terms are fractions instead of integers.

Vertex form is a(x -h)^2 +k. The coefficient of x inside parentheses is 1. The given form is not vertex form because the leading coefficient has not been removed to outside parentheses.

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2) What is the vertex of the parabola with the equation y=(x−2)^2+10?

A. (−2, −10)

B. (2, 10)

C. (−2, 10)

D. (2, −10)

Vertex form is a(x -h)^2 +k. Comparing to the given equation, we find the vertex (h, k) = (2, 10).

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3) For the given function, identify the x- and y-intercepts if any, the vertex, the axis of symmetry, and the maximum or minimum value. f(x)=−x^2+25

A. The x-intercepts are (−5,0) and (5,0). The y-intercept is (0,−25). The vertex is (0,−25). The axis of symmetry is x=0. The minimum value of the function is −25.

B. There are no x-intercepts. The y-intercept is (0,25). The vertex is (0,25). The axis of symmetry is y=0. The maximum value of the function is 25.

C. The x-intercepts are (−5,0) and (5,0). The y-intercept is (0,25). The vertex is (0,25). The axis of symmetry is x=0. The maximum value of the function is 25.

D. The x-intercepts are (−25,0) and (25,0). The y-intercept is (0,5). The vertex is (0,5). The axis of symmetry is x=0. The maximum value of the function is 5.

The x-intercepts are the values of x that make y=0. They are (±5, 0). The y-intercept is the value of y when x=0. It is (0, 25).

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4) A student says that the function f(x)=−x^2−9 has the x-intercepts (−3,0) and (3,0). Is the student correct? If not, explain why.

A. The student is correct.

B. The student is not correct. The equation f(x)=0 has one real solution, so the x-intercept is (9,0).

C. The student is not correct. The equation f(x)=0 does not have any real solutions, so the graph has only one x-intercept, (0,0).

D. The student is not correct. The equation f(x)=0 does not have any real solutions, so the graph does not have any x-intercepts.

The parabola opens downward and has a maximum value of -9, so cannot cross the x-axis. There are no x-intercepts, hence no real solutions.