Write the following expression in vertex form by completing the square
.𝑦 = 𝑥2 − 12𝑥 + 7
To write the expression in vertex form, we need to complete the square by adding and subtracting a constant term inside the parentheses.
First, we need to factor out the coefficient of 𝑥2, which is 1 in this case:
𝑦 = 𝑥2 − 12𝑥 + 7
= (𝑥 − 6)2 − 36 + 7 (since (a-b)^2 = a^2 -2ab + b^2, we can add and subtract (6)^2=36 inside the parentheses)
= (𝑥 − 6)2 − 29
Therefore, the vertex form of the expression 𝑦 = 𝑥2 − 12𝑥 + 7 is 𝑦 = (𝑥 − 6)2 − 29 . The vertex is at (6, -29).
y = x^2 - 12 x + 7
y = x^2 - 12 x + 6^2 ... - 6^2 + 7
y = x^2 - 12 x + 36 - 29
y = (x-6)^2 - 29
therefore symmetric about the veritcal line x = 6
and hits that line at y = -29
Yes, that's correct! Great job!
To rewrite the given quadratic equation in vertex form by completing the square, follow these steps:
Step 1: Group the quadratic terms together.
𝑦 = 𝑥^2 − 12𝑥 + 7
Step 2: Take out the common factor from the quadratic terms (𝑥^2 and −12𝑥).
𝑦 = (𝑥^2 − 12𝑥) + 7
Step 3: Calculate half of the coefficient of the linear term (-12/2) and square it.
Half of −12 is −6, and −6 squared is 36.
𝑦 = (𝑥^2 − 12𝑥 + 36) − 36 + 7
Step 4: Rewrite the equation by adding and subtracting the value obtained in Step 3 inside the parentheses.
𝑦 = (𝑥^2 − 12𝑥 + 36) − 29
Step 5: Factor the perfect square trinomial inside the parentheses.
The perfect square trinomial is (𝑥 − 6)^2, since (𝑥 − 6)(𝑥 − 6) = 𝑥^2 − 12𝑥 + 36.
𝑦 = (𝑥 − 6)^2 − 29
So, the equation 𝑦 = 𝑥^2 − 12𝑥 + 7 can be rewritten in vertex form as 𝑦 = (𝑥 − 6)^2 − 29.