Question 629298: Find the quadratic polynomial whose graph goes through the points (−1,5), (0,5), and (2,29).
f(x)= _x2+_ x+_ <<< that form
I got to the step 4a+2b+0=29
that was like my 6/7th step. I am getting stuck from there. Can i get any help please?
Found 2 solutions by ewatrrr, Theo:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi
quadratic polynomial whose graph goes through the points (-1,5), (0,5), and (2,29).
f(x) = ax^2 + bx + c
f(0) = 5 = c
f(x) = ax^2 + bx + 5 ****
a(-1)^2 + b(-1) + 5 = 5 || using (-1,5)
a*2^2 + 2b + 5 = 29 || using (2,29)
a - b = 0
4a +2b = 24 ||multiplying 1st EQ thru by 2 and adding it to the second
6a = 24
a = 4 and b = 4 
f(x) = 4x^2 + 4x + 5
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! it's a quadratic polynomial and it goes through the points:
(-1,5)
(0,5)
(2,29)
the general form of a quadratic equation is:
ax^2 + bx + c = 0
another form of the quadratic equation is:
y = ax^2 + bx + c
if it goes through the point (-1,5), then this equation becomes:
5 = a(-1)^2 + b(-1) + c
simplify this to get:
5 = a - b + c
if it goes through the point (0,5), then this equation becomes:
5 = c
if it goes through the point (2,29), then this equation becomes:
29 = a(2)^2 + b(2) + c
simplify this to get:
29 = 4a + 2b + c
we have 3 equations:
they are:
5 = a - b + c
5 = c
29 = 4a + 2b + c
since c = 5, we can substitute in the other 2 equations to get:
5 = a - b + 5
29 = 4a + 2b + 5
these simplify to:
a - b = 0
4a + 2b = 24
a - b = 0 results in a = b
substitute a for b in the other equation to get:
4a + 2a = 24
simplify to get:
6a = 24
a = 4
we now know that a = 4 and c = 5
we also know that b = which results in b = 4
we should have:
a = 4
b = 4
c = 5
let's see if that works.
our equation is:
y = 4x^2 + 4x + 5
when x is equal to -1, we get:
y = 4(-1)^2 + 4(-1) + 5 which simplifies to:
y = 4 - 4 + 5 which results in y = 5 which is correct because we are dealing with the point (-1,5)
our equation is:
y = 4x^2 + 4x + 5
when x = 0, our equation becomes:
y = 0 + 0 + 5 which results in y = 5 which is correct because we are dealing with the point (0,5)
when x = 2, our equations becomes:
y = 4(2)^2 + 4(2) + 5 which simplifies to:
y = 16 + 8 + 5 which simplifies further to:
y = 29 which is correct because we are dealing with the point (2,29)
looks like we're good and the equation i:
y = 4x^2 + 4x + 5
|
|
|
