Question

graph the parabola and give its vertex, axis of symmetry, x-intercepts, and y-intercept.

$y = -3x^2 + 12x - 6$

the vertex is $(2,6)$.
(type an ordered pair.)

the axis of symmetry is $x=2$.
(type an equation. use integers or fractions for any numbers in the equation.)

select the correct choice below and fill in any answer boxes within your choice.

a. the x-intercepts are at $x = \square$.
(type an exact answer, using radicals as needed. use a comma to separate answers as needed.)

b. there is no x-intercept.

Explanation:

Step1: Find x-intercepts

To find the x-intercepts, set \( y = 0 \) in the equation \( y=-3x^{2}+12x - 6 \). So we have the quadratic equation:

$$ -3x^{2}+12x - 6=0 $$

Divide the entire equation by \(-3\) to simplify:

$$ x^{2}-4x + 2=0 $$

Step2: Solve the quadratic equation

Use the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 1 \), \( b=-4 \), and \( c = 2 \).
First, calculate the discriminant \( \Delta=b^{2}-4ac=(-4)^{2}-4\times1\times2=16 - 8 = 8 \).
Then, find the roots:

$$ x=\frac{4\pm\sqrt{8}}{2}=\frac{4\pm2\sqrt{2}}{2}=2\pm\sqrt{2} $$

So the x - intercepts exist, and we choose option A.

Answer:

A. The x - intercepts are at \( x = 2+\sqrt{2},2-\sqrt{2} \)