Question

I need help with this too.. sketch the graph for f(x ) =x2-8x+16?

Answer 1

are you familiar with differential calculus?

Answer 2

this is a parabola to start off with

Answer 3

f(x ) =x2-8x+16
You can either plot points, or convert the equation into vertex form to and find the y intercepts of the function.
Vertex form: y=(x-h)^2+k
Vertex point is (h,k)
The first step to convert a strandard quadratic equation into vertex form is by completeing the square. Are you familiar with that?

Answer 4

it is an upward parabola,complete the square to find vertex

Answer 5

Yes, all quadratic equations are parabolas in graphs.

Answer 6

no wrote it out please?

Answer 7

work

Answer 8

First, do you know what is the vertex point on the parabola?

Answer 9

Or, how to find the x and y intercept of the function?

Answer 10

@Zale101 are you sure he's have to apply a parabola equation from analytic geometry?
I believe it's unnecesaryc complex.

Answer 11

no

Answer 12

How to complete the square
f(x) =x^2-8x+16
\[y=x^2-8x+16\]
Step 1: subtract 16 to both sides
\[y-16=(x^2-8x)\]
Step 2:Take the half of the coefficent, which is 8, and then square it

Answer 13

\((\frac{ 8 }{ 2 })^{2}\)=?

Answer 14

8

Answer 15

(8/2) is 4, you square it
4^2=16 :) so 8 was the wrong answer?

Answer 16

@owlcoffee i believe this question is algebra based. @keyshia.hart1 is that right?

Answer 17

Anyways, back to your problem. So, we got for (8/2)=4 and 4^2=16, you agree @keyshia.hart1 ?

Answer 18

\[(x^2-8x+16)=y-16-16\]
Now, factor (\(x^2-8x+16)\) what would you get?

Answer 19

http://www.purplemath.com/modules/sqrquad.htm
If you have no idea what i'm talking about, maybe this would help

Answer 20

im lost

Answer 21

In order to graph a function, we will have to first study the domain of the function, in other words, the values of x that can cause an indetermination on the function.
Since it's a quadratic function, we can be sure that no value of x will cause an indetermination on the domain, we write it like this:
\[d(f)= \mathbb{R} \]
Second, let's analyze the "roots" of the function, and later on, the y-interception.
Let's first begin with the roots of the function, we will call "root" to all those values of x that make the function equal zero, in other words, the x-axis interception.
so:
\[x^2 -8x+16=0\]
this is a quadratic equation, so we'll use the general formula to find the x values that satify it:
\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
\[x=\frac{ 8 \pm \sqrt{8^{2}-4(1)(16)} }{ 2(1) }\]
\[x=\frac{ 8 \pm \sqrt{0} }{ 2 }\]
\[x=4\]
since we got a square root of zero, means that the function will be tangent to the x-axis.
Now, since we got the roots, we can study the sign of the function, with a line that represents the values of the image of the x axis.
Since the squared component of the function is positive, I can be certain that the images to the right (x-values go bigger) the value of the function will remain positive:

it's positive on both sides, because we had a double root.
A double root is a special case, and indicates a point of tangency to the x-axis.
So, we can discard that the function would ever take negative values.
let's now find the "y-interception".
The y interception is the image of the function when x=0.
therefore:
\[f(0)=(0)^2-8(0)+16\]
\[f(0)=16\]
now, with all the information found, we can now sketch it:

(just imagine that it's well drawn)