Question

10 convert the vertex - form quadratic equation y = 3(x - 2)^2+4 to standard form (y = ax^2+bx + c) a y = 3x^2-8 b x = 8 + y-19x c y = 3x^2-12x + 16 d y=-x^2+5x - 4

Explanation:

Step1: Expand the square term

Use the formula $(a - b)^2=a^{2}-2ab + b^{2}$. Here $a = x$ and $b = 2$, so $(x - 2)^2=x^{2}-4x + 4$. Then $y=3(x - 2)^2+4=3(x^{2}-4x + 4)+4$.

Step2: Distribute the 3

Multiply 3 by each term inside the parentheses: $3(x^{2}-4x + 4)=3x^{2}-12x+12$. So $y=3x^{2}-12x + 12+4$.

Step3: Combine like - terms

Add the constant terms 12 and 4: $y=3x^{2}-12x+16$.

Answer:

C. $y = 3x^{2}-12x + 16$