Log in ☭СССР☭ 10 years agoPosted 10 years ago. Direct link to ☭СССР☭'s post “How do you convert a "ver...” How do you convert a "vertex form" equation into "standard form" equation? • (25 votes) Alex Tran 10 years agoPosted 10 years ago. Direct link to Alex Tran's post “y = a(x-h)^2 + k is the v...” y = a(x-h)^2 + k is the vertex form equation. Now expand the square and simplify. (26 votes) Anna Koen 9 years agoPosted 9 years ago. Direct link to Anna Koen's post “Where did you get the two...” Where did you get the two in the (2,5)? • (15 votes) NIKHIL SINGH 3 years agoPosted 3 years ago. Direct link to NIKHIL SINGH's post “It took me 15 minutes to ...” It took me 15 minutes to understand this but ,, >> (17 votes) acarulli21st 5 years agoPosted 5 years ago. Direct link to acarulli21st's post “in the graph why did you ...” in the graph why did you use 1 and 3 as random numbers? • (18 votes) Saivishnu Tulugu 5 years agoPosted 5 years ago. Direct link to Saivishnu Tulugu's post “You could have used whate...” You could have used whatever you wanted, but 1 and 3 were convenient as they produced whole numbers when plugged into the equation. (13 votes) Leone Alvarez 6 years agoPosted 6 years ago. Direct link to Leone Alvarez 's post “What if the a value is a ...” What if the a value is a fraction? How would I simply graph two coordinates that aren’t the vertex? • (16 votes) Levin Ma 4 years agoPosted 4 years ago. Direct link to Levin Ma's post “It is not possible to gra...” It is not possible to graph a parabola with only 2 points that are not a vertex, unless you also have the average rate of change or both points are on the same side of the parabola. It would be extremely difficult to graph it, and with most points impossible. For Q1, which value are you talking about? (3 votes) Ashna 5 years agoPosted 5 years ago. Direct link to Ashna's post “how would we find the y i...” how would we find the y intercept of an equation in vertex form? • (7 votes) Kim Seidel 5 years agoPosted 5 years ago. Direct link to Kim Seidel's post “For any equation, you can...” For any equation, you can find the y-intercept by using x=0 in the equation. Hope this helps. (12 votes) Mohamed Ibrahim 5 years agoPosted 5 years ago. Direct link to Mohamed Ibrahim's post “We can't make a fully acc...” We can't make a fully accurate parabola by hand, could we ? • (5 votes) Pyroaura 5 years agoPosted 5 years ago. Direct link to Pyroaura's post “Well, mathematically spea...” Well, mathematically speaking, there is no way to draw a "perfect" parabola by hand. Think of it like drawing a "perfect" circle. (12 votes) charlespmartin59 9 years agoPosted 9 years ago. Direct link to charlespmartin59's post “What if A is a fraction? ...” What if A is a fraction? Like; -1/3(x+2)^2 + 3 • (7 votes) Patrick 8 years agoPosted 8 years ago. Direct link to Patrick's post “Well, the best way to gra...” Well, the best way to graph this is to first convert this into standard form. We expand the square and combine any like terms: (8 votes) twentyonellamas 8 years agoPosted 8 years ago. Direct link to twentyonellamas's post “At 1:25, how does he dete...” At • (5 votes) Vu 8 years agoPosted 8 years ago. Direct link to Vu's post “y=a(x-h)²+kI already tol...” y=a(x-h)²+k If a is positive, the parabola open up and will have a minimum point at the vertex. y=-10(x+4)²-3 I encourage you try to graph y=a(x-h)²+k with different values it vill help you with a better understanding. You can go here and try your graphs. (10 votes) abkin2406 4 years agoPosted 4 years ago. Direct link to abkin2406's post “What if the first number ...” What if the first number is a fraction • (7 votes) Pranav Devanand 4 years agoPosted 4 years ago. Direct link to Pranav Devanand's post “It does not change the pr...” It does not change the process if the first number is a fraction. (5 votes) Hadassah17 9 years agoPosted 9 years ago. Direct link to Hadassah17's post “Ok. So why does the equat...” Ok. So why does the equation y= -2(x-2)^2+5 have a vertex (h,k) of (2,5) if the value of h is -2? • (7 votes) andrewp18 9 years agoPosted 9 years ago. Direct link to andrewp18's post “That is because the verte...” That is because the vertex form is: (5 votes)Want to join the conversation?
You should get y = a(x^2 -2hx + h^2) + k.
Multiply by the coefficient of a and get y = ax^2 -2ahx +ah^2 + k.
This is standard form of a quadratic equation, with the normal a, b and c in ax^2 + bx + c equaling a, -2ah and ah^2 + k, respectively.
as you can see , just substitute the value by this
a(x-h)^2 + k
-2(x-2)^2 + 5
Here Sal has taken (h,k) h = 2 and k = 5
don't confuse with the minus sign (-), if there is (-) between x and h thus the value will be positive i.e 2 .If the sign between x and h is positive then the value for 'h' in (h,k) will be negative . The value of the (h,k) is the vertex of parabola .
Don't worry you will get to know it if you analyse for 5 minutes .
You should watch it 3 to 4 times .
For example: y=2(x+3)^2-4
Set x=0: y=2(0+3)^2-4 = 2(9)-4 = 18-4 = 14
So the y-intercept is at: (0, 14)
f(x) = -1/3(x+2)^2 + 3
f(x) = -1/3(x+2)(x+2) + 3
f(x) = (-1/3 x - 2/3)(x+2) + 3
Use some paper and pencil for the next steps, like what I did:
f(x) = -1/3 x^2 - 4/3 x + 5/3
For starters, we can find the vertex first. Let's find the axis of symmetry:
x = (--4/3)/(2(-1/3))
x = -2
Now we plug -2 into the formula, and f(x) = 3. Graph (-2,3), and now, we use the quadratic formula to find the zeroes, which are -5 and 1. We graph the points on the x-axis and connect the three points with a curve. We're done!
1:25
Is k always the y term, and is the value of x that makes ax^2 = 0 the y value? I didn't quite get how he came to that.
I already told you about k which is the vertical shift and (x-h)² which is the horizontal shift on your other question. Now I will tell you about a.
If a is negative, the parabola open down and will have a maximum point at the vertex.
If |a| > 1 , it will have a "skinny" parabola comparing to y=x²
If |a| < 1, it will have a "fat" parabola comparing to y=x²
Try to graph y=x² (|a| =1) and then y=2x² (|a| > 1) and y=(1/2)x² ( |a| < 1) you will what I mean about skinny and fat parabola.
About the vertex, the vertex is determined by (x-h)² and k. The x value that makes x-h=0 will be the x-coordinate of the vertex. K will be the y-coordinate of the vertex.
This will have a vertex at (-4,-3). a is negative therefore will open down and have a maximum point. |-10| > 1 therefore it's really skinny.
https://www.desmos.com/calculator
y = a(x - h)² + k
In this case a = -2
k = 5
h must be 2 because the original form states -h. Therefore, we have:
-h = -2
h = 2
Comment if you have questions.
Video transcript
We're asked to graph theequation y is equal to negative 2 times x minus2 squared plus 5. So let me get by scratch pad outso we could think about this. So y is equal to negative 2times x minus 2 squared plus 5. So one thing, when you see aquadratic or a parabola graph expressed in this way, thething that might jump out at you is that this termright over here is always going to bepositive because it's some quantity squared. Or I should say, it's alwaysgoing to be non-negative. It could be equal to 0. So it's always going tobe some quantity squared. And then we're multiplyingit by a negative. So this whole quantityright over here is going to be non positive. It's always going to beless than or equal to 0. So this thing is alwaysless than or equal to 0, the maximum valuethat y will take on is when this thingactually does equal 0. So the maximumvalue for y is at 5. The maximum value for y is 5. And when does that happen? Well, y hits 5 whenthis whole thing is 0. And when does thisthing equal 0? Well, this whole thing equals0 when x minus 2 is equal to 0. And x minus 2 is equal to0 when x is equal to 2. So the point 2 comma5 is the maximum point for this parabola. And it is actuallygoing to be the vertex. So if we were to graph this,so the point 2 comma 5. So that's my y-axis. This is my x-axis. So this is 1, 2, 1, 2, 3, 4, 5. So this right here isthe point 2 comma 5. This is a maximum point,it's a maximum point for this parabola. And now I want to find two morepoints so that I can really determine the parabola. Three points completelydetermine a parabola. So that's 1, the vertex,that's interesting. Now, what I'd liketo do is just get two points that areequidistant from the vertex. And the easiest way to dothat is to maybe figure out what happens when x is equalto 1 and when x is equal to 3. So I could make a table hereactually, let me do that. So I care about x being equalto 1, 2, and 3, and what the corresponding y is. We already know that when x isequal to 2, y is equal to 5. 2 comma 5 is our vertex. When x is equal to 1, 1minus 2 is negative 1, squared is just 1. So this thing is goingto be negative 2 plus 5, so it's going to be 3. And when x is equal to3, this is 3 minus 2, which is 1 squared is 1 timesnegative 2 is negative 2 plus 5 is 3 as well. So we have three points. We have the point 1 comma3, the point 2 comma 5, and the point 3 comma3 for this parabola. So let me go backto the exercise and actually putthose three points in. And so we have the point 1 comma3, we have the point 2 comma 5, and we have the point 3 comma 3. And we have now fullydetermined our parabola.
