Questions: The equation of the parabola y=3x^2-6x+7 in vertex form is a.) y=3(x-1)^2+4 b.) y=3(x+1)^2+4 c.) y=3(x+1)^2-4 d.) y=3(x-1)^2-4

Transcript text: The equation of the parabola $y=3 x^{2}-6 x+7$ in vertex form is $\qquad$ a.) $y=3(x-1)^{2}+4$ b.) $y=3(x+1)^{2}+4$ c.) $y=3(x+1)^{2}-4$ d.) $y=3(x-1)^{2}-4$

Solution

Solution Steps

Step 1: Find the vertex $(h, k)$ of the parabola

To find the vertex of the parabola, we calculate $h = -\frac{b}{2a} = -\frac{-6}{2_3} = 1$ and $k = c - \frac{b^2}{4a} = 7 - \frac{-6^2}{4_3} = 4$. This gives us the coordinates of the vertex $(h, k)$.

Step 2: Write the equation in vertex form

Substituting $h = 1$ and $k = 4$ into the vertex form equation $y=a(x-h)^2+k$, we get: $$y = 3(x - (1))^2 + 4$$