Algebra

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.

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  1. Can you please show your work and let us know where you are getting stuck? Thanks!

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    Leo
  2. #1. You need to learn how to type fractions online. I assume you meant (3x+1/3)^2+8/9
    If so, then it's because the vertex form for a parabola is
    y = a(x-h)^2 + k
    So the correct choice is C

    #2. Given the vertex form shown above, the vertex is at (h,k)

    #3. set x=0 to get the y-intercept
    set y=0 to get the x-intercept(s)
    The vertex lies on the axis of symmetry, and will be either the min or max, depending on the coefficient of x^2

    #4.
    incorrect. The discriminant is negative
    Note that y = -(x^2+9)

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    oobleck

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