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Can someone please help me with this? I just can't seem to get the right answer ugh 

 

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

 Sep 1, 2017

Best Answer 

 #3
avatar+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
avatar+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
avatar+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
avatar+2345 
+1
Best Answer

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
avatar
+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

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