Question

rewrite this function in equivalent forms to identify th
$f(x) = 2x^2 - 4x - 6$
type the correct answer in each box.

key features:
y-intercept: (0, \square)
axis of symmetry: $x = \square$
vertex: (\square, \square)
x-intercepts: (-1, 0) and (\square, 0)

Explanation:

Step1: Find y - intercept

To find the y - intercept, we set \(x = 0\) in the function \(f(x)=2x^{2}-4x - 6\).
Substitute \(x = 0\) into the function: \(f(0)=2(0)^{2}-4(0)-6=-6\). So the y - intercept is \((0,-6)\).

Step2: Find axis of symmetry

For a quadratic function in the form \(f(x)=ax^{2}+bx + c\), the formula for the axis of symmetry is \(x=-\frac{b}{2a}\).
In the function \(f(x)=2x^{2}-4x - 6\), \(a = 2\) and \(b=-4\).
Substitute \(a = 2\) and \(b=-4\) into the formula: \(x=-\frac{-4}{2\times2}=\frac{4}{4} = 1\). So the axis of symmetry is \(x = 1\).

Step3: Find vertex

The x - coordinate of the vertex is the same as the axis of symmetry, so \(x = 1\).
To find the y - coordinate of the vertex, substitute \(x = 1\) into the function \(f(x)=2x^{2}-4x - 6\).
\(f(1)=2(1)^{2}-4(1)-6=2 - 4-6=-8\). So the vertex is \((1,-8)\).

Step4: Find the other x - intercept

We know that for a quadratic function \(f(x)=ax^{2}+bx + c\), if the roots are \(x_1\) and \(x_2\), then \(x_1\times x_2=\frac{c}{a}\) (from Vieta's formulas).
In the function \(f(x)=2x^{2}-4x - 6\), \(a = 2\), \(c=-6\), and one root \(x_1=-1\). Let the other root be \(x_2\).
We have \(-1\times x_2=\frac{-6}{2}=-3\), so \(x_2 = 3\). So the other x - intercept is \((3,0)\).

Answer:

y - intercept: \((0,\boxed{-6})\)
axis of symmetry: \(x=\boxed{1}\)
vertex: \((\boxed{1},\boxed{-8})\)
x - intercepts: \((-1,0)\) and \((\boxed{3},0)\)