+99 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.
+2446 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!
+1904 \(y=(x+h)2\)
\(y=2(x+h)\)
\(y=2x+2h\)
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 \(y=2x-4\)
+99 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?
+2446 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!
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
