How do you solve this since there isn't a c? I need the vertex points (ie: (2,5)) I'm in the 8th grade (Algebra 1) and our teacher neglected to tell us how to do these.

If there is no C, it is understood to be 0 (zero).

Good luck!

How do you solve this since there isn't a c? I need the vertex points (ie: (2,5)) I'm in the 8th grade (Algebra 1) and our teacher neglected to tell us how to do these.

Tutors, please sign in to answer this question.

If there is no C, it is understood to be 0 (zero).

Good luck!

When a quadratic function like this one is set equal to 0 (meaning it is in its general form), you are typically looking to solve for the roots/zeros of the function.

Since you are looking to solve for the vertex of the graph of the quadratic function, which is a parabola, it needs to be in vertex form. If it's not in vertex form, you need to convert it to this form from the standard form of the function.

Quadratic function in standard form looks like the following:

y = ax^{2} + bx + c , where a ≠ 0

The vertex form of the function looks like the following:

y = a(x - h)^{2} + k , where a ≠ 0 and (h, k) is the vertex of the parabola

For the problem in question, the function needs to be set equal to y in order to continue with converting it into vertex form and find the vertex.

y = 9k^{2} + 45k

Factor out the leading coefficient from the right hand side of the equation:

y = 9(k^{2} + 5k)

Now complete the square for the equation inside the parentheses by taking half of b (the coefficient of x) and squaring it. That is, (b/2)^{2} = (5/2)^{2} = 5^{2}/2^{2} = 25/4

Add and subtract this value inside the parenthesis so as to not change the equation:

y = 9(k^{2} + 5k + 25/4 - 25/4)

Notice that the first 3 terms in the parenthesis now make a perfect square trinomial, which can be factored and yields the following:

y = 9((k^{2} + 5k + 25/4) - 25/4)

y = 9((k + 5/2)(k + 5/2) - 25/4))

y = 9((k + 5/2)^{2} - 25/4)

Distributing the 9 into each term inside the parentheses, we get:

y = 9·(k + 5/2)^{2} - 9·(25/4)

y = 9(k + 5/2)^{2} - 225/4

Looking back at the vertex form of a quadratic, the sign inside the parenthesis needs to be negative and the sign outside the parenthesis needs to be positive. Thus,

y = 9(k - (-5/2))^{2} + (-225/4)

With the function in its vertex form and given that the vertex is at (h, k), then since

h = -5/2 and k = -225/4

the vertex of the parabola is at (-5/2, -225/4)

Nikunj P.

Math and Science Specialist, Over 14 Years of Educational Experience

Scarsdale, NY

4.9
(88 ratings)

Evan F.

Mathematics graduate looking to enhance your mathematics ability.

Richmond Hill, NY

4.9
(201 ratings)

John E.

Physics and Math tutor - Masters from Cornell and Columbia University

New York, NY

4.8
(124 ratings)