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.

CrazyDaizy Sep 1, 2017

#3**+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

#1**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 \(y=2x-4\)

.gibsonj338 Sep 1, 2017

#2**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**+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**+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

Guest Sep 2, 2017