Question

Find the vertex, x/y intercept(s), domain, range,

axis of symmetry, and end behavior of the
quadratic function f(x) = 2x^2-
8x + 4


can someone please help.. i have to record a video of me answering it

Answer 1

See below for answers and explanations (along with attached graph)

Explanation:

It best helps to convert the equation into vertex form because it provides a lot of information about the characteristics of the parabola:

`f(x)=2x^2-8x+4\\\\f(x)=2(x^2-4x+2)\\\\f(x)+2(2)=2(x^2-4x+2+2)\\\\f(x)+4=2(x^2-4x+4)\\\\f(x)+4=2(x-2)^2\\\\f(x)=2(x-2)^2-4`

Now, recall that vertex form is
`y=a(x-h)^2+k` where
`(h,k)` is the vertex, the axis of symmetry is the line
`x=h`, and
`a` is the growth or shrink factor.

Vertex

This is pretty simple as you can just look at the equation, so
`(h,k)\rightarrow(2,-4)`

Axis of Symmetry

Again, also simple, so
`x=2` would be the line.

X-intercepts

The beauty of having converted the equation to vertex form is that when we set
`f(x)=0`, we can easily find our x-intercepts with minimal work:

`f(x)=2(x-2)^2-4\\\\0=2(x-2)^2-4\\\\4=2(x-2)^2\\\\2=(x-2)^2\\\\\pm√(2)=x-2\\\\x=2\pm√(2)`

So, our x-intercepts are at the points
`(2+√(2),0)` and
`(2-√(2),0)`.

Y-intercept

This is just a simple substitution of
`x=0` and everything works out nicely:

`f(x)=2(x-2)^2-4\\\\f(0)=2(0-2)^2-4\\\\f(0)=2(-2)^2-4\\\\f(0)=2(4)-4\\\\f(0)=8-4\\\\f(0)=4`

So, our y-intercept would be at the point
`(0,4)`.

Domain

If you try any real number for
`x`, there will always be a value for
`f(x)`, so the domain of the function is
`(-\infty,\infty)` in interval notation.

Range

Recall back to when we found our vertex of
`(2,-4)`. Because the leading coefficient of the function is positive, our vertex is the minimum, which means
`f(x)` cannot be lower than -4, but anything higher works, so our range is
`[-4,\infty)` in interval notation.

End Behavior

We can easily see that if we plug in a large value for
`x`, then
`f(x)` will also be a large value, so our end behavior for the function is that as
`x\rightarrow\infty`, then
`f(x)\rightarrow \infty`. Another good indicator of this end behavior is the even leading degree of 2.

I hope these explanations helped! Please feel free to view the attached graph below to help give you a visual!