To find the [tex]\(x\)[/tex]-intercepts of the parabola defined by the equation [tex]\( y = 2x^2 - 8x + 6 \)[/tex], we need to determine the points where the graph of the equation crosses the [tex]\(x\)[/tex]-axis. These are the points where [tex]\( y = 0 \)[/tex].
Given the quadratic equation:
[tex]\[
y = 2x^2 - 8x + 6
\][/tex]
Setting [tex]\( y = 0 \)[/tex], we have:
[tex]\[
0 = 2x^2 - 8x + 6
\][/tex]
We need to solve this quadratic equation for [tex]\( x \)[/tex]. The general form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex]. Here, [tex]\( a = 2 \)[/tex], [tex]\( b = -8 \)[/tex], and [tex]\( c = 6 \)[/tex].
To find the [tex]\(x\)[/tex]-intercepts, we use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
First, calculate the discriminant:
[tex]\[
\text{Discriminant} = b^2 - 4ac = (-8)^2 - 4(2)(6)
\][/tex]
Simplify the expression inside the square root:
[tex]\[
\text{Discriminant} = 64 - 48 = 16
\][/tex]
Now we can compute the two potential solutions for [tex]\( x \)[/tex]:
[tex]\[
x_1 = \frac{-(-8) + \sqrt{16}}{2(2)} = \frac{8 + 4}{4} = \frac{12}{4} = 3
\][/tex]
[tex]\[
x_2 = \frac{-(-8) - \sqrt{16}}{2(2)} = \frac{8 - 4}{4} = \frac{4}{4} = 1
\][/tex]
These solutions give us the [tex]\(x\)[/tex]-coordinates of the intercepts. Since [tex]\( y = 0 \)[/tex] at the [tex]\(x\)[/tex]-intercepts, the corresponding ordered pairs are:
[tex]\[
(3, 0) \quad \text{and} \quad (1, 0)
\][/tex]
Hence, the [tex]\(x\)[/tex]-intercepts of the parabola [tex]\( y = 2x^2 - 8x + 6 \)[/tex] are:
[tex]\[
(3.0, 0), (1.0, 0)
\][/tex]
