Question

The graph of a quadratic function passes through the points (-1,0), (-2,0), and (0,2). Write the quadratic function in general form.

Answer 1

Answer:

$f(x)=x^2+3x+2$

Step-by-step explanation:

step 1

we have the points

(-1,0), (-2,0), and (0,2)

Plot the points

using a graphing tool

see the attached figure

The graph of a quadratic function must be a vertical parabola open upward

The vertex is a minimum

The quadratic function in general form is equal to

$f(x)=ax^2+bx+c$

Substitute the value of x and the value of y of each given ordered pair in the general equation and solve for a,b and c

(0,2)

For x=0, y=2

substitute

$2=a(0)^2+b(0)+c$

$c=2$

(-1,0)

For x=-1, y=0

substitute

$0=a(-1)^2+b(-1)+2$

$0=a-b+2$ ----> $a=b-2$ ----> equation A

(-2,0)

For x=-2, y=0

substitute

$0=a(-2)^2+b(-2)+2$

$0=4a-2b+2$ ----> equation B

we have the system

$a=b-2$ ----> equation A

$0=4a-2b+2$ ----> equation B

substitute equation A in equation B

$0=4(b-2)-2b+2$

solve for b

$0=4b-8-2b+2$

$2b=6$

$b=3$

Find the value of a

$a=3-2=1$

therefore

The quadratic function in general form is equal to

$f(x)=x^2+3x+2$

see the attached figure N 2 to better understand the problem