A quadratic function that fits the points (0,5) (2,3) (4,9)

Step-by-step explanation:

y=ax^2+bx+c we are given point (0,5) so if x=0, y=c, so c=5. We have two other points

(2,3)(4,9) which gives us

4a+2b+5=3 and 16a+4b+5=9

Subtracting twice the first from the second

16a+4b+5-8a-4b-10=9-6

8a-5=3

8a=8, so a=1, so using 4a+2b+5=3 we get

4(1)+2b+5=3

2b+9=3

2b=-6

b=-3 so our quadratic is

y=x^2-3x+5

incomplete question. how do you expect anyone to you when you do not post the question in its entirety?

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?