Question

Please help me with "Graphs of quadratic functions" please. I don't understand anything about this stuff.

Answer 1

Do you want to learn how to graph quadratic function?

Answer 2

quadratic means; of a square, from the old guys studying quadralaterals... the name simply suggests that the highest exponent value is 2

Answer 3

a quad is one step up from a line, and so it bends in one place; the place it bends is called its vertex

Answer 4

es I wanna learn hoe to graph them. how to find the vertex, etc.

Answer 5

ax^2 +b x +c

Vertex= -b/2a

Answer 6

the equation of a quad is defined primarily from its vertex

Answer 7

(x-h)^2+k
In this case vertex is (h,k)

Answer 8

y = M(x-h)^2+k; where h and k are the x and y components of the vertex

Answer 9

what is the equation for finding the verex?

Answer 10

by alot of slacking off and trying to avoid too much work; the old guys developed formulas for quads .... the most useful one is in 2 parts;

-b/2a and +- sqrt(b^2-4ac)/2a

Answer 11

the -b/2a is the vertexes x component

Answer 12

the part under the sqrt is what tells us if we can actually play nice with the equation

Answer 13

irregardless, and other misappropriated words later, the determinate (-b/2a) will always give us the x component for the vertex

Answer 14

how do you find the y

Answer 15

so how do we locate these a b cs you may wonder ...

from the form: ax^2 +bx +c

Answer 16

you find y simply by inserting the value of the determinate into the equation and solving

Answer 17

so say:
y=x^2-6x+6
(3,-3) would be the vertex?

Answer 18

lets see; what is our determinate? our -b/2a?

b = -6 and a = 1 therefore; --6/2 = 3 for our x value, insert this into the equation to find y

y = (3)^2 -6(3) +6
= 9 -18 +6
= -9 + 6
= -3

so we find that; (3,-3) is the vertex point

Answer 19

Thank you. Now if we graph that, is the vertex all i need to do so?

Answer 20

the vertex simply tells you where it bends at..... in order to graph it, we can look at what the equation is telling us.

the first term is positive; that means it opens upwards; and we see that the vertex is below the x axis telling us that it cross when y = 0

to find those zeros we can use the whole quad formula; the first part we already found.

x = 3 +- sqrt(6^2 -4(6))/2
= 3 +- sqrt(36 -24)/2
= 3 +- sqrt(12)/2
= 3 +- 2sqrt(3)/2
= 3 +- sqrt(3)

Answer 21

the place it crosses the y axis is the last number in the equation: (0,6)

all these points need to be accounted for in the graph to get it looking right, but to get it looking real good, you need to table up alot of points

Answer 22

but if youre just trying to pick out the graph of a given equation; use the clues inside the equation :)

Answer 23

thank you, this helped me alot.:) I do appreciate it dearly, have a good day.