How do you find the x-coordinate of the vertex for the graph 4x2+16x+12=0

Question

How do you find the x-coordinate of the vertex for the graph  4x2+16x+12=0

nebajcioz · Accepted Answer

Step 1
Your quadratic equations is in standard form

a x^{2} + b x + c

so the x-coordinate of the vertex (also called the axis of symmetry) is found with the formula:

a o s = \frac{- b}{2 a}

4 x^{2} + 15 x + 12

a o s = \frac{- 16}{2 \times 4 = - 2}

Graph:

{4 x^{2} + 16 + 12 [- 10, 10, - 5, 5]}

shimmertulipsog · Accepted Answer

Step 1 Simplifying 4x2+16x+12=0 Reorder the terms: 12+16x+4x2=0 Solving 12+16x+4x2=0 Solving for variable &#039;x&#039; Factor out the Greatest Common Factor (GCF), &#039;4&#039; 4(3+4x+x2)=0 Factor a trinomial. 4((3+x)(1+x))=0 Ignore the factor 4. Step 2 Set the factor &#039;(3+x)&#039; equal to zero and attempt to solve: Simplifying 3+x=0 Solving 3+x=0 Move all terms containing x to the left, all other terms to the right. Add &#039;-3&#039; to each side of the equation. 3±3+x=0±3 Combine like terms: 3±3=0 0+x=0pn3 x=0±3 Combine like terms: 0±3=−3 x=−3 Simplifying x=−3 Step 3 Set the factor &#039;(1+x)&#039; equal to zero and attempt to solve: Simplifying 1+x=0 Solving 1+x=0 Move all terms containing x to the left, all other terms to the right. Add &#039;-1&#039; to each side of the equation. 1±1+x=0±1 Combine like terms: 1±1=0 0+x=0±1 x=0±1 Combine like terms: 0±1=−1 x=−1 x={−3, −1}

How do you find the x-coordinate of the vertex for

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