How do I put this into Vertex form?
y=6x^2+7x+5
To put the given quadratic equation into vertex form, we need to complete the square. The vertex form of a quadratic equation is given by:
y = a(x - h)² + k
where (h, k) represents the coordinates of the vertex.
Let's start by identifying the values of a, b, and c in the given equation. In the equation y = 6x^2 + 7x + 5, the coefficient of x^2 is 6 (which represents a), the coefficient of x is 7 (which represents b), and the constant term is 5 (which represents c).
Step 1: Divide the coefficient of x^2 by 2 and square it. This will give us the value we need to complete the square. In this case, (7/2)^2 = 49/4.
Step 2: Add the value obtained in Step 1 inside the parentheses and subtract it outside the parentheses. Rearrange the equation accordingly:
y = 6(x^2 + (7/2)x + 49/4 - 49/4) + 5
Step 3: Simplify the equation inside the parentheses:
y = 6[(x + 7/2)^2 - 49/4] + 5
Step 4: Distribute and combine like terms:
y = 6(x + 7/2)^2 - 6(49/4) + 5
= 6(x + 7/2)^2 - 147/2 + 5
Step 5: Simplify further to get the equation in vertex form:
y = 6(x + 7/2)^2 - 137/2
Therefore, the given quadratic equation y = 6x^2 + 7x + 5 in vertex form is y = 6(x + 7/2)^2 - 137/2.
What method have you learned?
There are several.