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.

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

#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)

1) The correct answer is A. The expression has a constant outside of the squared term. In the vertex form of a quadratic function, the squared term should have a coefficient of 1. In this case, the squared term has a coefficient of 9, not 1.

2) The vertex of the parabola with the equation y=(x−2)2+10 is B. (2, 10). In the vertex form y=a(x-h)2+k, the vertex is represented by the point (h, k). In this case, the vertex is (2, 10).

3) The correct answer is 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. To find the x-intercepts, we set y=0 and solve for x. The y-intercept is found by setting x=0. The vertex can be found using the formula x = -b/2a. The axis of symmetry is the vertical line passing through the vertex. The maximum or minimum value can be determined from the y-coordinate of the vertex, which is -25 in this case.

4) The student is correct. The function f(x)=−x2−9 does have the x-intercepts (-3,0) and (3,0). To find the x-intercepts, we set y=0 and solve for x. In this case, when we substitute y=0 into the equation and solve for x, we get x=-3 and x=3. Therefore, the x-intercepts are (-3,0) and (3,0).

1) To determine whether f(x)=(3x+13)2+89 is the vertex form of f(x)=9x^2+2x+1, we need to compare the two expressions. The vertex form of a parabola is given by f(x) = a(x-h)^2 + k, where (h, k) represents the vertex of the parabola.

In the given expression f(x)=(3x+13)2+89, there is a constant term (89) outside the squared term, which is not present in the equation f(x)=9x^2+2x+1. This means that the given expression is not in the vertex form.

Therefore, the correct option is A. The expression has a constant outside of the squared term.

2) The vertex form of a parabola is given by f(x) = a(x-h)^2 + k, where (h, k) represents the vertex of the parabola. From the equation y=(x−2)^2+10, we can see that the vertex form of the parabola is y = (x-2)^2 + 10.

Comparing this with the standard vertex form, we can identify that the vertex of the parabola is (2, 10).

Therefore, the correct option is B. (2, 10).

3) To find the x-intercepts of a function, we need to solve the equation f(x) = 0. Similarly, to find the y-intercept, we substitute x = 0 into the function and solve for y.

For the function f(x) = -x^2 + 25, we can find the x-intercepts by setting f(x) = 0:

-x^2 + 25 = 0

This equation can be rearranged as x^2 = 25, which gives us x = -5 and x = 5. So the x-intercepts are (-5, 0) and (5, 0).

To find the y-intercept, we substitute x = 0 into the function:

f(0) = -(0^2) + 25 = -25

So the y-intercept is (0, -25).

By comparing the given options, we can see that the correct option is 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.

4) To determine whether the function f(x) = -x^2 - 9 has the x-intercepts (-3, 0) and (3, 0), we need to solve the equation f(x) = 0:

-x^2 - 9 = 0

This equation can be rearranged as x^2 = -9, which has no real solutions. Therefore, the graph of this function does not intersect the x-axis, and it has no x-intercepts.

Hence, the correct option is 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.