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Vertex Form
This page will teach you everything needed to know about the vertex form equation of a parabola.
Vertex form= a(x-h)^2+k(Look back for reference if needed)
Axis of Symmetry and Optimal Value
So we are going to start off with the simpler side of figuring out how to graph a parabola when given the vertex form equation and that is figuring out the axis of symmetry and optimal value, so let's get started. Before getting started however you must understand what exactly is the axis of symmetry and optimal value. The axis of symmetry is the h value in the parabola and decides how many units the parabola moves horizontally, we will go into further detail afterwards. The optimal value is the y value in the parabola and decides how many units the parabola moves vertically. Oh and one more thing, the variable x=axis of symmetry and the variable y=optimal value just to clear any possible confusion. To simplify x=h and y=k.
The axis of symmetry is basically the h value in the equation(x=h).
For example if I have this equation: y=(x-3)^2+5 the axis of symmetry would be 3, now you may be asking "why a positive 3 if it's negative in the equation?" Bare in mind that the sign is always flipped when looking for the axis of symmetry in a vertex form equation(i.e +3 becomes -3, -3 becomes +3). So therefore so far we know that the parabola will move 3 units to the right on the x-axis and that's pretty much what the axis of symmety covers.
Now for the optimal value it's pretty much the same thing as the axis of symmetry except now we're dealing with the y-axis and not the x-axis, so here we go.
The optimal value is basically the k value in the equation(y=k)
We will use the same equation as before "y=(x-3)^2+5". The optimal value is 5, simple as that, the sign does not flip for the k value ever, only for the h value it flips, it's as simple as that. This basically means that the parabola will be moving 5 units up, if the 5 was negative then it would be 5 units down(i.e "y=(x-3)^2-5").
Terminology
-Translation: To move an object to a certain distance away from it's current location, it is not a translation however if that certain object is rotated, flipped or re-sized.
Vertex
As you can see this is what the graph of the equation y=(x-3)^2+5 looks like. For now only focus on the axis of symmetry and optimal value, notice how the parabola is 3 units to the right and 5 units up. We will get into the actual curve and everything else afterwards.
We are now dealing with the vertex. The vertex is the highest or lowest point in a parabola. If a parabola opens upwards it means it has a minimum vertex(lowest point), if it opens downwards it has a maximum vertex(highest point). The way you find the vertex from a vertex form equation is very simple. Find the axis of symmetry(x=h) and optimal value(y=k), then put both of them in brackets like this, "(h, k)", those coordinates are your vertex. The axis of symmetry always comes first then the optimal value respectively, so once you have your vertex simply plot it on the graph and your parabola will be based off of that point.
-Upwards parabola
-It has a minimum vertex
-Vertex=(0,0)
-Downwards parabola
-It has a maximum vertex
-Vertex=(0,0)
Reflection
The reflection in a parabola is which way the parabola is reflected. Is it facing downwards or upwards? The way you would figure out which way the parabola is reflected is by looking at the "a" value in the vertex form equation. If the a value is positive the parabola opens upwards, if it is negative the parabola opens downwards, i.e if we use the same equation as before
y=(x-3)^2+5 we can tell it opens upwards, because the a value is an imaginary positive 1 as there is no a value to be seen, however if we look at the same equation but this time I add a negative in the a value's place, y=-(x-3)^2+5 it opens downwards, because now the a value is a negative 1. Remember if there is no number in the a value's place it is a "1".
Zeroes(x-intercepts)
The zeroes in a parabola are the x-intercepts of it. There are either 0, 1, or 2 zeroes in a parabola. How do you find out how many zeroes there are? It is very simple, look at the k value in the vertex form equation, if k is greater than 0 there are no x-intercepts. If k is less than 0 there are 2 x-intercepts. If k is equal to 0(k=0), then there is 1 x-intercept. The way you solve for the x-intercepts with the vertex form equation is like this:
We will work with the equation "y=(x-3)^2-5". First look at k, it is -5, less than 0, so there are 2 x-intercepts.
Step 1- Make y=0, so...
0=(x-3)^2-5
Step 2- Move y over to the right side, so...
(x-3)^2-5=0
Step 3- Move k over to the right, so...
(x-3)^2=5
Step 4- Move the a value over to the right and put it beneath 5(divide 5 by "a", or make a the denominator), so...
(x-3)^2=5/1
Step 5- We now have to get rid of "^2", since we're squaring the equation we move it to the right and square root the quotient, so...
5/1=5
(x-3)=√5
Step 6- Solve the square root, so...
√5=2.2360679775
Step 7- You will need to memorize this symbol, "±", this symbol is known as the plus minus sign, how you use it is like this, i.e if I do this 5±4 you would get 2 answers, because you're doing 5+4=9 and 5-4=1, that's how you use it. So what we do now is we move the -3 over to the right and ± 2.2360679775, so...
x=3±2.2360679775
x=5.236(approximately)
x=0.764(approximately)
Once you have your zeroes you simply plot them on your graph, remember there are no y values in a x-intercept coordinate. So the coordinates you would plot are (5.236, 0), (0.764, 0), there are your x-intercepts(zeroes).
Graph of y=(x-3)^2-5
x-intercepts
Vertex
Vertex
Vertical Stretch or Compression
The vertical stretch or compression in a parabola is how narrow or wide it is respectively. The way you know whether a parabola is stretched or compressed is by looking at the "a" value, if a is greater than 1 the parabola is stretched, if a is less than 1 the parabola is compressed. The way you would figure out how narrow or wide the parabola is is by using the "step-pattern", which you will learn next.
Step-Pattern
The step pattern is the method used to solve the vertical stretch or compression of a parabola. Say we now are working with the equation "y=2(x-3)^2+5", notice "a" is now 2, instead of 1, this means the parabola is now narrower. How we find out how narrow it is, is by doing this...
Only focus on the 2, forget all other values, only look at a.
The step-pattern is 1, 3, 5, 7, etc. Starting from 1 and increasing by 2 each time. You can go up in the step-pattern as far as you would like.
Step 1- Multiply each value in the step-pattern by the a value, in our case 2, so...
1 x 2=2
3 x 2=6
5 x 2=10
7 x 2=14
Step 2- Now starting from your vertex, wherever that might be, move 1 unit to the right, and 2 units up, 1 unit to the right, 6 units up, 1 unit to the right, 10 units up, 1 unit to the right, 14 units up, do the same for the left side as well. Once you are done plotting all those points, connect the dots.
Now if we work with the same equation but change a to 0.5(1/2, so "y=1/2(x-3)^2+5") the parabola will be compressed as a is < 1, so doing the same thing as before...
1 x 1/2=0.5
3 x 1/2=1.5
5 x 1/2=2.5
7 x 1/2=3.5
We now have solved the step-pattern. What you do now is, starting from your vertex, move one unit right, 0.5 units up, one unit right, 1.5 units up, one unit right, 2.5 units up, 1 unit right, 3.5 units up, do the same for the left side as well. Connect the dots.
Remember no matter what the a value is, be it 0.2, 0.5, 2, 3, 4, 5, even 1000 the same process applies.
The green parabola has no stretch or compression, made using the equation, "y=(x-3)^2+5
The red parabola is stretched, made using the equation, "y=2(x-3)^2+5"
The blue parabola is compressed, made using the equation, "y=1/2(x-3)^2+5"
If you are still having trouble with this section, watch this video: https://www.youtube.com/watch?v=SFc4F0iUF3U
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