write the equation of the parabola and a vertex form and standard form given the following information

see explanation

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

(9)

Here (h, k) = (3, - 2), thus

y = a(x - 3)² - 2

To find a substitute (2, 3) into the equation

3 = a(2 - 3)² - 2 ( add 2 to both sides )

5 = a(- 1)² = a , then

y = 5(x - 3)² - 2 ← in vertex form

y = 5(x² - 6x + 9) - 2

   = 5x² - 30x + 45 - 2

y = 5x² - 30x + 43 ← in standard form

(10)

Here (h, k) = (2, - 5), thus

y = a(x - 2)² - 5

To find a substitute (5, 4) into the equation

4 = a(5 - 2)² - 5 ( add 5 to both sides )

9 = a(- 3)² = 9a ( divide both sides by 9 )

a = 1

y = (x - 2)² - 5 ← in vertex form

y = x² - 4x + 4 - 5

y = x² - 4x - 1 ← in standard form

 9) y = 5(x - 3)² - 2         y = 5x² - 30x + 43

              10) y = (x - 2)² - 5           y =  x² - 4x + 2

Step-by-step explanation:

The Vertex form of a quadratic equation is:    y = a(x - h)² + k     where

The Standard form of a quadratic equation is: y = ax² + bx + c

--> this is found by expanding the vertex form and adding like terms.

Input Vertex (h, k) and Point (x, y) into the Vertex form to solve for "a"

9) Vertex (h, k) = (3, -2) and Point (x, y) = (2, 3)

   3 = a(2 - 3)² - 2

  3 = a(-1)² - 2

  5 = a(1)

  5 = a

Vertex form: y = 5(x - 3)² - 2

Standard Form: y = 5(x² - 6x + 9) - 2

                          y = 5x² - 30x + 45 - 2

                          y = 5x² - 30x + 43

10) Vertex (h, k) = (2, -5) and Point (x, y) = (5, 4)

  4 = a(5 - 2)² - 5

  4 = a(3)² - 5

  9 = a(9)

  1 = a

Vertex form: y = (x - 2)² - 5

Standard Form: y = (x² - 4x + 4) - 2

                          y =  x² - 4x + 2

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!