^{2}+ bx + c(2.5)

^{2}+k

The coordinates of the x-intercepts given are (-1,0) and (3,0). The y-intercept is 2.5. How do you create an equation in standard and vertex form with that info?

I'm not sure what each letter represents except that standard forms' "c" refers to the y-intercept

ax^{2} + bx + c(2.5)

a(x-h)^{2} +k

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Here’s an alternate method that highlights some of the basic concepts of quadratic functions.

The x-values of the x-intercepts are also the zeros of f(x). So we can write: f(x) = a (x - (-1)) (x - 3) = a (x + 1) (x - 3)

We can find the value of a by substituting the y-intercept, (0, 5/2):

The x-values of the x-intercepts are also the zeros of f(x). So we can write: f(x) = a (x - (-1)) (x - 3) = a (x + 1) (x - 3)

We can find the value of a by substituting the y-intercept, (0, 5/2):

f(0) = 5/2 = a (0 + 1) ( 0 - 3) = -3a; so a = -5/6 and:

f(x) = -5/6 (x + 1) (x - 3), which is the Factored Form.

Expanding the Factored Form:

f(x) = -5/6 (x^2 - 2x - 3),

f(x) = (-5/6) x^2 + (5/3) x + 5/2, which is the Standard Form.

Note that the Axis of Symmetry is a vertical line half way between the zeros, so its equation is: x = (-1 + 3)/2 = 1.

The vertex point, (h,k), is the intersection of the Axis of Symmetry and the function, so (h,k) = (1,f(1)):

k = f(1) = -5/6 (1 + 1) (1 - 3) = 10/3, and

f(x) = -5/6 (x - 1)^2 + 10/3 is the Vertex Form of the function.

f(x) = -5/6 (x + 1) (x - 3), which is the Factored Form.

Expanding the Factored Form:

f(x) = -5/6 (x^2 - 2x - 3),

f(x) = (-5/6) x^2 + (5/3) x + 5/2, which is the Standard Form.

Note that the Axis of Symmetry is a vertical line half way between the zeros, so its equation is: x = (-1 + 3)/2 = 1.

The vertex point, (h,k), is the intersection of the Axis of Symmetry and the function, so (h,k) = (1,f(1)):

k = f(1) = -5/6 (1 + 1) (1 - 3) = 10/3, and

f(x) = -5/6 (x - 1)^2 + 10/3 is the Vertex Form of the function.

As a check all three forms were graphed using GeoGebra and all three were the same graph with zeros at -1 and 3 and y-intercept at 2.5.

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

In your problem you should mention equation of quadratic.

f(x) = aX^{2} + bx +c

Y intercept is always = c , because f(0) = a(0)^{2} + b(0) + c = c

Every Quadratic of aX^{2} + bX + c, with roots (X intercept) of m.n equals

f(x) = k ( X- m)( X-n)

So : quadratic of the problem:

k( X +1) ( X-3) = k ( X^{2}- 2X -3 )

To find the value of K:

2.5 = k ( 0 ^{2}- 2( 0) - 3) / coordinate of Y intercept ( 0, 2.5)

k = 2.5/- 3 = - 5/6

f(x) =- 5/6 X^{2} + 5/3 X + 5/2

In the vertex form:

K ( X - b/ 2a) ^{2} ±f ( -b/2a)/ also X _{Vertex =}( 3 -1)/2 =1

-5/6 ( X - 1 ) ^{2 }+ f(1) =-5/6 ( X-1)^{2} + 10/3

If you are not familiar with functional notation :

f(1) = value of Y at X=1

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