For $f(x) = 2x^2 - 4x -1$ , what is the vertex and axis of symmetry?

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Answer 1 · 2017-12-09T20:22:41

vertex at $(1,-3)$ , the axis of symmetry is at $x=1$

There are two ways of doing this - completing the square or through differentiation. Either way, we are trying to find the minimum point.

Completing the square

$f(x)=2x^2-4x-1$
$=2(x^2-2x-1/2)$
$=2([x-1]^2-1-1/2)$
$=2([x-1]^2-3/2)$
$=2[x-1]^2-3$

This means the minimum value is at $f(x)=-3$ , where $x=1$ .

The axis of symmetry will be the same as the x-coordinate of the minimum, this being $x=1$

So the vertex is at $(1,-3)$ , the axis of symmetry is at $x=1$

Differentiation

I'd generally only differentiate if the number were a bit awkward, if completing the square would give loads of fractions.

$f(x)=2x^2-4x-1$
$f'(x)=4x-4$
Let $f'(x)=0$
$4x-4=0$
$x=1$
Let $x=1$
$f(1)=2(1)^2-4(1)-1$
$=-3$

So the vertex is at $(1,-3)$ , the axis of symmetry is at $x=1$

For f(x) = 2x^2 - 4x -1f(x) = 2x^2 - 4x -1, what is the vertex and axis of symmetry?