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 5 solutions by ewatrrr, Theo, ikleyn, n2, Edwin McCravy:
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
Answer by ikleyn(52958)  (Show Source):
You can put this solution on YOUR website! .
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?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This task, as it is solved in the post by @Theo, seems routine and does not create any inspiration.
But it depends on how to look at it.
I will show you another solution from different perspective that will make it seem beautiful like a sparkling diamond.
Notice that the y-coordinate is the same, '5', for both points (-1,5) and (0,5).
It means that if you consider another quadratic function, g(x) = f(x) - 5, instead of f(x),
this new quadratic function will have x-intercepts at x= -1 and x= 0.
In turn, it means that g(x) = ax*(x+1) with some real coefficient 'a',
so f(x) = ax(x+1) + 5.
Great (!) So, to determine f(x), we only need to find single unknown coefficient 'a'.
For it, we use the condition that the parabola passes through the point (2,29).
It gives us this equation
f(2) = 29, or a*2*(2+1) + 5 = 29, or 2*3*a = 29 - 5, or 6a = 24.
Thus we find a = 24/6 = 4.
So, our quadratic function is f(x) = 4x(x+1) + 5, or f(x) = 4x^2 + 4x + 5.
At this point, the problem is solved completely, and we obtained the solution
with minimal computations and maximal fun.
This trick works for many other similar problem, where
two points of the three given points have the same y-coordinate.
So, this method is worth to learn ( ! )
Answer by n2(8)  (Show Source):
You can put this solution on YOUR website! .
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?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Notice that the y-coordinate is the same, '5', for both points (-1,5) and (0,5).
It means that if you consider another quadratic function, g(x) = f(x) - 5, instead of f(x),
this new quadratic function will have x-intercepts at x= -1 and x= 0.
In turn, it means that g(x) = ax*(x+1) with some real coefficient 'a',
so f(x) = ax(x+1) + 5.
Great (!) So, to determine f(x), we only need to find single unknown coefficient 'a'.
For it, we use the condition that the parabola passes through the point (2,29).
It gives us this equation
f(2) = 29, or a*2*(2+1) + 5 = 29, or 2*3*a = 29 - 5, or 6a = 24.
Thus we find a = 24/6 = 4.
So, our quadratic function is f(x) = 4x(x+1) + 5, or f(x) = 4x^2 + 4x + 5.
At this point, the problem is solved completely, and we obtained the solution
with minimal computations and maximal fun.
This trick works for many other similar problem, where
two points of the three given points have the same y-coordinate.
So, this method is worth to learn ( ! )
Answer by Edwin McCravy(20067)  (Show Source):
You can put this solution on YOUR website!
On your TI-84
press STAT
press ENTER
Under L1 enter the x-coordinates and under L2 enter the y-coordinates.
That is, make the screen look like this under L1 and L2
L1 | L2
-1 | 5
0 | 5
2 | 29
Press STAT then right arrow once to highlight CALC
Press 5 to choose this--> 5: QuadReg
Press ENTER five times
You read
y=ax2+bx+c
a=4
b=4
c=5
The answer is f(x) = 4x2 + 4x + 5
Edwin
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