Need help with da mafs Pls get it right lol I already have an 80 on this assignment
1 answer:
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The vertex is located at (h,k)
The two roots p and q can be defined as such
- p = h-r
- q = h+r
where r is the horizontal distance between the vertex and either root. We use the same r value for each equation because of symmetry. The diagram below shows what I mean.
Let's add up those two equations to get
p+q = 2h
The left hand sides p and q add to p+q. The right hand sides add to (h-r)+(h+r) = 2h. Note the r's cancel.
From here, we divide both sides by 2 and we end up with
h = (p+q)/2
In short, if we know the roots p and q, then we find the x coordinate of the vertex (h) to be the average of those roots. You add up the roots and then divide by 2.
After you figure out the value of h, finding the value of k will have us plug in this h value into the quadratic equation and simplifying.
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Example:
Let's say we have the quadratic y = x^2 - 8x + 12. This factors to (x-2)(x-6)
Solving (x-2)(x-6) = 0 shows that x-2 = 0 becomes x = 2 and x-6=0 becomes x = 6.
So the roots here are p = 2 and q = 6. The order of the roots doesn't matter.
Averaging these roots gives us
h = (p+q)/2
h = (2+6)/2
h = 8/2
h = 4
The x coordinate of the vertex is x = 4.
Plug this back into the original equation to find...
y = x^2 - 8x + 12
y = (4)^2 - 8(4) + 12
y = 16 - 32 + 12
y = -16 + 12
y = -4
Or you could say
y = x^2 - 8x + 12
y = (x-2)(x-6)
y = (4-2)*(4-6)
y = (2)*(-2)
y = -4
Either way, we find that k = -4 is the y coordinate of the vertex
The vertex for this example is located at (h,k) = (4, -4)
Refer to the diagram below.
