Question

Determine the equation of the parabola passing through the points (-3,13), (0,1), and (1, -7)​

Answer 1

y=-5x+1

Explanation:

The equation is: y=mx+b

m = the slope of the line

b = the y-intercept (0,b)

The y-intercept is (0,1) so b = (0,1)

So the equation would be y =mx+1

Now in order to calculate the slope, the equation is:

y₂-y₁ over

x₂-x₁

So we should use the points (-3,13) and (1,-7).

-7-13 over (over means a fraction symbol)

1+3

Simplify:

-20 over 4 =-5/1 = -5

M=-5

So the equation is now:

y=-5x+1

Answer 2

Explanation:

You need to do some solving simultaneously to get these values. Your quadratic equation is of the form

`ax^2+bx+c=y`

Use the coordinates you've been given to solve 3 equations. It will be super simple if we start with the coordinate (0, 1). Here's why (obvious after some substitution is done):

`a(0)^2+b(0)+c=1` which gives us that

c = 1. Now we have a variable to plug in for c to solve for a and b. Again, we have coordinates that we can use to create 2 more equations:

`a(-3)^2+b(-3)+1=13` and, simplified:

9a - 3b = 12

and the second equation is:

`a(1)^2+b(1)+1=-7` and, simplified:

a + b = -8

Now combine the 2 bold equations and solve by elimination or substitution to find either a or b. I chose elimination and multiplied the second equation by 3 to get a new equation:

3a + 3b = -24

Using the elimination method:

9a - 3b = 12

3a + 3b = -24

You can see that the b's subtract each other away, leaving us with

12a = -12 so

a = -1

Now plug -1 in for a to solve for b:

-1 + b = -8 so

b = -7 and the quadratic is

`-x^2-7x+1=y`