The function shown is in the form y = (x - h)2. Determine the value of h.

The equation graphed in the picture is not linear (because the graph is not a straight line). It is a quadratic, however, because it forms a parabola. I will use the quadratic in "vertex" form. 

 

$f(x)=a(x-h)^2+k$ where $(h,k)$ is the vertex.

 

However, we do not know what the value of a is. Therefore, I will resort to the "standard" quadratic form:
 

$f(x)=ax^2+bx+c$

 

Before we can identify the equation that contains this parabola, let's see if we can determine the y-intercept. It would appear as if it is located at (0,4). Let's plug that in.

 

$f(x)=ax^2+bx+c$	Plug this in when x=0.
$f(0)=a(0)^2=b*0+c$	Of course, the 0's make simplification much easier.
$f(0)=c$	Now, we know what $f(0)$ equals because we know the corresponding y-coordinate when x=0. It happens to be 4.
$4=c$

 

Now, let's use another point. This time, I will use the vertex, which appears to be located at (2,0). Let's do the exact same process.

 

$f(x)=ax^2+bx+c$	Replace every instance of x with 2, in this case, as that is the x-coordinate we want.
$f(2)=a(2)^2+b*2+c$	Let's simplify the right hand side.
$f(2)=4a+2b+ c$	Of course, we know that $f(2)=0$.
$0=4a+2b+c$

 

We need 1 more point. For a quadratic, you always need 3 points to determine its equation. Because quadratic equations have symmetry along its axis of symmetry, we know that (4,4) is also a point on the parabola. Let's plug that in, and we will utilize the exact same process.

 

$f(4)=a(4)^2+b*4+c$	Yet again, simplify the right hand side.
$f(4)=16a+4b+c$	We know that $f(4)=4$, so plug that in.
$4=16a+4b+c$

 

Great! Let's look at these equations side-by-side:

 

{ $4=c$

{ $0=4a+2b+c$
{ $4=16a+4b+c$

 

Of course, if 4=c, then we can make this look a tad nicer:
 

{ $0=4a+2b+4$

{ $4=16a+4b+4$

 

Let's make it nicer again! To make it easier, the first equation has a GCF of 2, and the second equation has a GCF of 4. Simplifying this will make the numbers in our system of equations smaller and easier to work with.

 

{ $0=2a+b+2$
{ $1=4a+b+1$

 

Now, let's solve this system of equations! I will use the substitution method because it is easier to understand when I write. I'll use the first equation first. My goal here is to isolate a variable. Since b has a coefficient of 1 already, I think I will isolate that variable. 

 

$0=2a+b+2$	Subtract b on both sides.
$-b=2a+2$	Divide by -1 on both sides to isolate b.
$b=-2a-2$

 

Plug this into b for equation 2:

 

$1=4a+b+1$	Since $b=-2a-2$ according to our previous equation, plug that in.
$1=4a+(-2a-2)+1$	Simplify the right hand side of the equation. The parentheses aren't actually necessary, but it might make it clearer where the substitution is taking place.
$1=2a-1$	Add 1 to both sides.
$2=2a$	Divide by 2 on both sides to isolate a.
$a=1$

 

Great! We know that a=1. We actually need not solve for b, in this case, as we know what the vertex is already, and we can use vertex form. As a recap, "vertex" form is the following:

 

$f(x)=a(x-h)^2+k$ where (h,k) is the vertex

 

Well, a=1, and the vertex is (2,0). Plug that in.

 

$f(x)=1(x-2)^2+0$

 

This can be simplified.

 

$f(x)=(x-2)^2$

 

This means that a=1, h=2, and k=0. I hope this helps!

$y=(x+h)2$

 

$y=2(x+h)$

 

$y=2x+2h$

 

First find a point on the graph and replace $x$ and $y$ for the point.  The point I will use is the point $(2,0)$

 

$y=2x+2h$

 

$0=2(2)+2h$

 

Next solve for $h$.

 

$0=2(2)+2h$

 

$0=4+2h$

 

$0-4=4+2h-4$

 

$-4=4+2h-4$

 

$-4=2h-0$

 

$-4=2h$

 

$\frac{-4}{2}=\frac{2h}{2}$

 

$-\frac{4}{2}=\frac{2h}{2}$

 

$-\frac{2}{1}=\frac{2h}{2}$

 

$-2=\frac{2h}{2}$

 

$-2=\frac{1h}{1}$

 

$-2=1h$

 

$-2=h$

 

$h=-2$

 

The answer is $h=-2$.  The equation is $y=2(x+(-2))$ or $y=2(x-2)$ or $y=2x-4$

You are told that the equation of the graph is of the form $\displaystyle y=(x-h)^{2}$

Looking at the graph it would seem that

$\displaystyle y = 0 \text{ when }x=2,\text{ so, substituting in those values, }\\ 0=(2-h)^{2},\\\text{ so, }h=2.$ 

The equation is $\displaystyle y=(x-2)^{2}.$

 

Tiggsy