How do you convert quadratic equations from general and factored form to vertex form?
Vertex form is: y= a(x-h)^2 +k
And from the vertex form how do you find the vertex.
Example questions:
y= -3x^2 + 21x
y=2x^2-10x+5
y=5x^2+5x-3
y=-3x^2-24x+7
-b/2a will be the h and what you get after plugging -b/2a in will be your k
first example y=a(x-7/2)^2+36 3/4
second example y=2(x+5/4)^2+20 5/8
third example y=5(x+1/2)^2-4 1/4
fourth example y=-3(x+4)^2+55
Consider (Eg 1).
Take -3 out common.
=> -3( x^2 - 7x)
= -3( x^2 - 2(1)(7/2) + 49/4 - 49/4)
= -3[ (x-7/2)^2 - 49/4]
h and k are the vertices. h = 7/2 and k = 147/4.
find dy/dx from y = ax^2 + bx +c
which is =2ax +b
set dy/dx=0
so, 2ax+b=0
or, x = -b/2a
this gives the x coordinate of the vertex
put this x value in the original equation to find the y coordinate of the vertex
y =
a(x^2 + b/a x + (b/2a)^2) - a*(b/2a)^2 +c
a(x + b/2a)^2 - b^2/4a +c
y = -3(x² - 7x)
y = -3(x² - 7x + (7/2)² - (7/2)²)
y = -3(x² - 7x + (7/2)²) - (-3)(7/2)²
y = -3(x - 7/2)² + 147/4
h = 7/2, k = 147/4
y = 2x² - 10x + 5
y = 2(x² - 5x + 25/4) - 2(25/4) + 5
y = 2(x² - 5x + 25/4) - 50/4 + 5
y = 2(x - 5/2)² - 30/4
h = 5, k = -15/2
y = 5x² + 5x - 3
y = 5(x² + x + (1/2)²) - 5(1/4) - 3
y = 5(x + 1/2)² - 17/4
h = -1/2, k = -17/4
y = -3x² - 24x + 7
y = -3(x² + 8x + 16) - (-3)(16) + 7
y = -3(x + 4)² + 55
h = -4, k = 55
#If you have any other info about this subject , Please add it free.# |