"Assume that the graph of the polynomial p(x)=ax2+bx+cpasses through the point (−1,2)and has a horizontal tangent at (12,−9). Find the sum a+b+c." I've already substitute (-1, 2) into the polynomial equation to get a - b + c = 2 and differentiate the original polynomial to get 2ax+b=0 and replaced x with 12 to get 24a + b = 0 as well as utilizing (12, -9) to get 144a + 12 b + c = -9 but I have no idea where to go from there..
your steps are correct so far
you now have 3 equations with 3 unknowns
solve these to find a, b and c and then calculate a+b+c
I've tried to use row operations but was unsuccessful, is there a particular order that I should place these equations?
You could just use simple substitution, e.g. start by eliminating c using the last equation
I'm not quite sure I understand you, use 144a + 12b + c = -9 to eliminate c?
c = -9 -12b -144a from that equation now substitute this into your first equation
Oh, the a-b+c=2 equation?
yes :)
you will then be left with 2 equations in two unknowns - a and b
How does that help? I have -13b-143a=11 right now.
and you also have: 24a + b = 0
which gives you: b = -24a substitute that into: -13b-143a=11 and solve for a
ok what do you need help with, solving for a?
a would equal 2/169?
not quite - you must have made a simple algebraic mistake
Did you not get 312a-143a=2?
-13b-143a=11 -13(-24a) - 143 a = 11 312a - 143a = 11 169a = 11 /169 on both sides a = 11/169
remember the negative sign
me? It went away
I am talking to @Clarence
ok sorry just trying to help
So b = - 264/169?
yes :)
good job!!
now you are just left with c to calculate and then you can work out what a+b+c equals
c being 63/169?
yup :)
wow sounds like you got it down, now just add them together
So all together it'd be -190/169?
yes - well done! :)
correct awesome job
Oh I get it now, thanks!
why do I feel like there is an echo in here :)
Probably because there is ;) Thanks again guys/girls!
haha you keep typing and replying right before i do and i don't see it sorry
Aha, all good :)
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