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

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

Sep 1, 2017

#3
+2345
+1

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!

Sep 1, 2017

#1
+1904
0

$$y=(x+h)2$$

$$y=2(x+h)$$

$$y=2x+2h$$

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 $$y=2x-4$$

.
Sep 1, 2017
#2
+99
0

oh wow! No wonder I could not figure out what I was doing :/ One question though, my answer choices are 1, -1, 2 and -1/2

Could it be positive 2 instead of -2?

CrazyDaizy  Sep 1, 2017
#3
+2345
+1

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!

TheXSquaredFactor Sep 1, 2017
#4
+1

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

Sep 2, 2017