Question

give me step by step of vertex form for y=1/4x^2-4

144

likes720 views

Lurline

4.6

81 Answers

Start with the given equation:

y = \frac{1}{4}x^2 - 4

Step 1: Factor out the coefficient of x^2

y = \frac{1}{4}(x^2) - 4

Step 2: Recognize that x^2 is already a perfect square, and we don't need to adjust for any x term since there isn't one. So we add and subtract 0 inside the bracket:

y = \frac{1}{4}(x-0)^2 - 4

Step 3: This form resembles the vertex form of a quadratic equation y = a(x-h)^2 + k, where h and k are the coordinates of the vertex.

Hence, the vertex form is:

y = \frac{1}{4}(x - 0)^2 - 4

So, the vertex form of the given quadratic equation is:

y = \frac{1}{4}(x - 0)^2 - 4

Step 1: Factor out the coefficient of x^2

Step 2: Recognize that x^2 is already a perfect square, and we don't need to adjust for any x term since there isn't one. So we add and subtract 0 inside the bracket:

Step 3: This form resembles the vertex form of a quadratic equation y = a(x-h)^2 + k, where h and k are the coordinates of the vertex.

Hence, the vertex form is:

So, the vertex form of the given quadratic equation is:

Frequently asked questions (FAQs)

Math Question: What is the factored form of the quadratic equation "x^2 + 5x + 6"?

+

Question: What is the derivative of ∫(0 to x) t^3 + 2t - 1 dt according to the Fundamental Theorem of Calculus?

+

What is the solution to the inequality 5x - 7 > 18?

+

New questions in Mathematics