How to create quadratic equation given $y$ intercept, and maximum and $B=8$?












0












$begingroup$


The given are



Two x-intercepts



y-intercept(0,-4)



Maximum at (2,4)



i tried everything i know...its been a long time since I have been doing math problems but the only way i thought about was to use -b/2A but i can't go anywhere after that...if possible just a hint to solve it is ...or if possible anyone tell me how to solve it?? The only other way i can think is point slope formula....






calculus algebra-precalculus conic-sections quadratics







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$endgroup$








  • 1




    $begingroup$
    Are you familiar with the vertex, or shifted, form of writing the equation for a parabola, $y = a(x - h)^2 + k$?
    $endgroup$
    – pjs36
    Aug 25 '15 at 23:16
















0












$begingroup$


The given are



Two x-intercepts



y-intercept(0,-4)



Maximum at (2,4)



i tried everything i know...its been a long time since I have been doing math problems but the only way i thought about was to use -b/2A but i can't go anywhere after that...if possible just a hint to solve it is ...or if possible anyone tell me how to solve it?? The only other way i can think is point slope formula....






calculus algebra-precalculus conic-sections quadratics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are you familiar with the vertex, or shifted, form of writing the equation for a parabola, $y = a(x - h)^2 + k$?
    $endgroup$
    – pjs36
    Aug 25 '15 at 23:16














0












0








0





$begingroup$


The given are



Two x-intercepts



y-intercept(0,-4)



Maximum at (2,4)



i tried everything i know...its been a long time since I have been doing math problems but the only way i thought about was to use -b/2A but i can't go anywhere after that...if possible just a hint to solve it is ...or if possible anyone tell me how to solve it?? The only other way i can think is point slope formula....






calculus algebra-precalculus conic-sections quadratics







share|cite|improve this question











$endgroup$




The given are



Two x-intercepts



y-intercept(0,-4)



Maximum at (2,4)



i tried everything i know...its been a long time since I have been doing math problems but the only way i thought about was to use -b/2A but i can't go anywhere after that...if possible just a hint to solve it is ...or if possible anyone tell me how to solve it?? The only other way i can think is point slope formula....






calculus algebra-precalculus conic-sections quadratics




calculus algebra-precalculus conic-sections quadratics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 25 '15 at 23:24









Rory Daulton

29.5k63355




29.5k63355










asked Aug 25 '15 at 23:14









MATH ASKERMATH ASKER

3802522




3802522








  • 1




    $begingroup$
    Are you familiar with the vertex, or shifted, form of writing the equation for a parabola, $y = a(x - h)^2 + k$?
    $endgroup$
    – pjs36
    Aug 25 '15 at 23:16














  • 1




    $begingroup$
    Are you familiar with the vertex, or shifted, form of writing the equation for a parabola, $y = a(x - h)^2 + k$?
    $endgroup$
    – pjs36
    Aug 25 '15 at 23:16








1




1




$begingroup$
Are you familiar with the vertex, or shifted, form of writing the equation for a parabola, $y = a(x - h)^2 + k$?
$endgroup$
– pjs36
Aug 25 '15 at 23:16




$begingroup$
Are you familiar with the vertex, or shifted, form of writing the equation for a parabola, $y = a(x - h)^2 + k$?
$endgroup$
– pjs36
Aug 25 '15 at 23:16










1 Answer
1






active

oldest

votes


















1












$begingroup$

You can start by writing out the general form of a quadratic polynomial: $P(x)=ax^2+bx+c$. Saying that the y-intercept is $(0,-4)$ is the same as saying that $P(0)=-4$, which gives us $c=-4$. Now the derivative of the polynomial will be $P'(x)=2ax+b$. Since the derivative of a differentiable function at a maximum is 0, you must have $P'(2)=0$ or $4a=-b$. Since the point $(2,4)$ is assumed to be in the graph of the function, we know that $P(2)=4$ or $4a+2b-4=4$, which means that $4a=8-2b$. Thus we have



$$4a=-b ;;;;text {and};;; 4a=8-2b$$



Which gives $b=8$ and $a=-2$, so we get $P(x)=-2x^2+8x-4$



EDIT



Another way to do it, as pjs36 said, would be to use the fact that every quadratic equation is of the form $P(x)=alpha(x-beta)^2 +gamma$ for some $alpha,; beta,; gamma$. When the equation is written in this form it becomes clear that $alpha$ must be negative (otherwise the function won't have a maximum), and then the maximum is attained when $x-beta =0$. Since the maximum is at $x=2$, we have $beta =2$. Also, the value of $P$ at the maximum will be $4=P(2)=alpha (0)^2 + gamma = gamma$, so $gamma =4$. Lastly, we know that the y-intercept is $(0,-4)$, so that $-4=P(0)=alpha(0-beta )^2 + gamma = 4alpha +4$, which gives $alpha=-2$. Now we can expand the original expression and we get $P(x)=alpha (x- beta)^2 +gamma =-2(x-2)^2 + 4=-2x^2 +8x -4$, so we get the same result as before.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much for the answer, may i ask how you got p'(x) = 2ax + b or just 2? if i get what you wrote, P(2)= 4 bc of the x square right, than how did you get 2b??
    $endgroup$
    – MATH ASKER
    Aug 26 '15 at 0:31










  • $begingroup$
    Sure! I got the that the derivative of P(x) is 2ax+b through some rules of differentiation: the derivative of $ax^2$ is $2xa$, the derivative of $bx$ is $b$, and the derivative of $c$ is $0$ because $c$ is a constant. Thus the derivative of the whole polynomial is $2ax+b+0$. $P(2)=4$ was part of the information you gave, because you said that there is a maximum at $(2,4)$, which means that at $x=2$, we must have $y=4=P(2)$.
    $endgroup$
    – Aitor Ormazabal
    Aug 26 '15 at 8:15











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1 Answer
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$begingroup$

You can start by writing out the general form of a quadratic polynomial: $P(x)=ax^2+bx+c$. Saying that the y-intercept is $(0,-4)$ is the same as saying that $P(0)=-4$, which gives us $c=-4$. Now the derivative of the polynomial will be $P'(x)=2ax+b$. Since the derivative of a differentiable function at a maximum is 0, you must have $P'(2)=0$ or $4a=-b$. Since the point $(2,4)$ is assumed to be in the graph of the function, we know that $P(2)=4$ or $4a+2b-4=4$, which means that $4a=8-2b$. Thus we have



$$4a=-b ;;;;text {and};;; 4a=8-2b$$



Which gives $b=8$ and $a=-2$, so we get $P(x)=-2x^2+8x-4$



EDIT



Another way to do it, as pjs36 said, would be to use the fact that every quadratic equation is of the form $P(x)=alpha(x-beta)^2 +gamma$ for some $alpha,; beta,; gamma$. When the equation is written in this form it becomes clear that $alpha$ must be negative (otherwise the function won't have a maximum), and then the maximum is attained when $x-beta =0$. Since the maximum is at $x=2$, we have $beta =2$. Also, the value of $P$ at the maximum will be $4=P(2)=alpha (0)^2 + gamma = gamma$, so $gamma =4$. Lastly, we know that the y-intercept is $(0,-4)$, so that $-4=P(0)=alpha(0-beta )^2 + gamma = 4alpha +4$, which gives $alpha=-2$. Now we can expand the original expression and we get $P(x)=alpha (x- beta)^2 +gamma =-2(x-2)^2 + 4=-2x^2 +8x -4$, so we get the same result as before.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much for the answer, may i ask how you got p'(x) = 2ax + b or just 2? if i get what you wrote, P(2)= 4 bc of the x square right, than how did you get 2b??
    $endgroup$
    – MATH ASKER
    Aug 26 '15 at 0:31










  • $begingroup$
    Sure! I got the that the derivative of P(x) is 2ax+b through some rules of differentiation: the derivative of $ax^2$ is $2xa$, the derivative of $bx$ is $b$, and the derivative of $c$ is $0$ because $c$ is a constant. Thus the derivative of the whole polynomial is $2ax+b+0$. $P(2)=4$ was part of the information you gave, because you said that there is a maximum at $(2,4)$, which means that at $x=2$, we must have $y=4=P(2)$.
    $endgroup$
    – Aitor Ormazabal
    Aug 26 '15 at 8:15
















1












$begingroup$

You can start by writing out the general form of a quadratic polynomial: $P(x)=ax^2+bx+c$. Saying that the y-intercept is $(0,-4)$ is the same as saying that $P(0)=-4$, which gives us $c=-4$. Now the derivative of the polynomial will be $P'(x)=2ax+b$. Since the derivative of a differentiable function at a maximum is 0, you must have $P'(2)=0$ or $4a=-b$. Since the point $(2,4)$ is assumed to be in the graph of the function, we know that $P(2)=4$ or $4a+2b-4=4$, which means that $4a=8-2b$. Thus we have



$$4a=-b ;;;;text {and};;; 4a=8-2b$$



Which gives $b=8$ and $a=-2$, so we get $P(x)=-2x^2+8x-4$



EDIT



Another way to do it, as pjs36 said, would be to use the fact that every quadratic equation is of the form $P(x)=alpha(x-beta)^2 +gamma$ for some $alpha,; beta,; gamma$. When the equation is written in this form it becomes clear that $alpha$ must be negative (otherwise the function won't have a maximum), and then the maximum is attained when $x-beta =0$. Since the maximum is at $x=2$, we have $beta =2$. Also, the value of $P$ at the maximum will be $4=P(2)=alpha (0)^2 + gamma = gamma$, so $gamma =4$. Lastly, we know that the y-intercept is $(0,-4)$, so that $-4=P(0)=alpha(0-beta )^2 + gamma = 4alpha +4$, which gives $alpha=-2$. Now we can expand the original expression and we get $P(x)=alpha (x- beta)^2 +gamma =-2(x-2)^2 + 4=-2x^2 +8x -4$, so we get the same result as before.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much for the answer, may i ask how you got p'(x) = 2ax + b or just 2? if i get what you wrote, P(2)= 4 bc of the x square right, than how did you get 2b??
    $endgroup$
    – MATH ASKER
    Aug 26 '15 at 0:31










  • $begingroup$
    Sure! I got the that the derivative of P(x) is 2ax+b through some rules of differentiation: the derivative of $ax^2$ is $2xa$, the derivative of $bx$ is $b$, and the derivative of $c$ is $0$ because $c$ is a constant. Thus the derivative of the whole polynomial is $2ax+b+0$. $P(2)=4$ was part of the information you gave, because you said that there is a maximum at $(2,4)$, which means that at $x=2$, we must have $y=4=P(2)$.
    $endgroup$
    – Aitor Ormazabal
    Aug 26 '15 at 8:15














1












1








1





$begingroup$

You can start by writing out the general form of a quadratic polynomial: $P(x)=ax^2+bx+c$. Saying that the y-intercept is $(0,-4)$ is the same as saying that $P(0)=-4$, which gives us $c=-4$. Now the derivative of the polynomial will be $P'(x)=2ax+b$. Since the derivative of a differentiable function at a maximum is 0, you must have $P'(2)=0$ or $4a=-b$. Since the point $(2,4)$ is assumed to be in the graph of the function, we know that $P(2)=4$ or $4a+2b-4=4$, which means that $4a=8-2b$. Thus we have



$$4a=-b ;;;;text {and};;; 4a=8-2b$$



Which gives $b=8$ and $a=-2$, so we get $P(x)=-2x^2+8x-4$



EDIT



Another way to do it, as pjs36 said, would be to use the fact that every quadratic equation is of the form $P(x)=alpha(x-beta)^2 +gamma$ for some $alpha,; beta,; gamma$. When the equation is written in this form it becomes clear that $alpha$ must be negative (otherwise the function won't have a maximum), and then the maximum is attained when $x-beta =0$. Since the maximum is at $x=2$, we have $beta =2$. Also, the value of $P$ at the maximum will be $4=P(2)=alpha (0)^2 + gamma = gamma$, so $gamma =4$. Lastly, we know that the y-intercept is $(0,-4)$, so that $-4=P(0)=alpha(0-beta )^2 + gamma = 4alpha +4$, which gives $alpha=-2$. Now we can expand the original expression and we get $P(x)=alpha (x- beta)^2 +gamma =-2(x-2)^2 + 4=-2x^2 +8x -4$, so we get the same result as before.






share|cite|improve this answer











$endgroup$



You can start by writing out the general form of a quadratic polynomial: $P(x)=ax^2+bx+c$. Saying that the y-intercept is $(0,-4)$ is the same as saying that $P(0)=-4$, which gives us $c=-4$. Now the derivative of the polynomial will be $P'(x)=2ax+b$. Since the derivative of a differentiable function at a maximum is 0, you must have $P'(2)=0$ or $4a=-b$. Since the point $(2,4)$ is assumed to be in the graph of the function, we know that $P(2)=4$ or $4a+2b-4=4$, which means that $4a=8-2b$. Thus we have



$$4a=-b ;;;;text {and};;; 4a=8-2b$$



Which gives $b=8$ and $a=-2$, so we get $P(x)=-2x^2+8x-4$



EDIT



Another way to do it, as pjs36 said, would be to use the fact that every quadratic equation is of the form $P(x)=alpha(x-beta)^2 +gamma$ for some $alpha,; beta,; gamma$. When the equation is written in this form it becomes clear that $alpha$ must be negative (otherwise the function won't have a maximum), and then the maximum is attained when $x-beta =0$. Since the maximum is at $x=2$, we have $beta =2$. Also, the value of $P$ at the maximum will be $4=P(2)=alpha (0)^2 + gamma = gamma$, so $gamma =4$. Lastly, we know that the y-intercept is $(0,-4)$, so that $-4=P(0)=alpha(0-beta )^2 + gamma = 4alpha +4$, which gives $alpha=-2$. Now we can expand the original expression and we get $P(x)=alpha (x- beta)^2 +gamma =-2(x-2)^2 + 4=-2x^2 +8x -4$, so we get the same result as before.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 26 '15 at 8:44

























answered Aug 25 '15 at 23:23









Aitor OrmazabalAitor Ormazabal

55129




55129












  • $begingroup$
    Thank you very much for the answer, may i ask how you got p'(x) = 2ax + b or just 2? if i get what you wrote, P(2)= 4 bc of the x square right, than how did you get 2b??
    $endgroup$
    – MATH ASKER
    Aug 26 '15 at 0:31










  • $begingroup$
    Sure! I got the that the derivative of P(x) is 2ax+b through some rules of differentiation: the derivative of $ax^2$ is $2xa$, the derivative of $bx$ is $b$, and the derivative of $c$ is $0$ because $c$ is a constant. Thus the derivative of the whole polynomial is $2ax+b+0$. $P(2)=4$ was part of the information you gave, because you said that there is a maximum at $(2,4)$, which means that at $x=2$, we must have $y=4=P(2)$.
    $endgroup$
    – Aitor Ormazabal
    Aug 26 '15 at 8:15


















  • $begingroup$
    Thank you very much for the answer, may i ask how you got p'(x) = 2ax + b or just 2? if i get what you wrote, P(2)= 4 bc of the x square right, than how did you get 2b??
    $endgroup$
    – MATH ASKER
    Aug 26 '15 at 0:31










  • $begingroup$
    Sure! I got the that the derivative of P(x) is 2ax+b through some rules of differentiation: the derivative of $ax^2$ is $2xa$, the derivative of $bx$ is $b$, and the derivative of $c$ is $0$ because $c$ is a constant. Thus the derivative of the whole polynomial is $2ax+b+0$. $P(2)=4$ was part of the information you gave, because you said that there is a maximum at $(2,4)$, which means that at $x=2$, we must have $y=4=P(2)$.
    $endgroup$
    – Aitor Ormazabal
    Aug 26 '15 at 8:15
















$begingroup$
Thank you very much for the answer, may i ask how you got p'(x) = 2ax + b or just 2? if i get what you wrote, P(2)= 4 bc of the x square right, than how did you get 2b??
$endgroup$
– MATH ASKER
Aug 26 '15 at 0:31




$begingroup$
Thank you very much for the answer, may i ask how you got p'(x) = 2ax + b or just 2? if i get what you wrote, P(2)= 4 bc of the x square right, than how did you get 2b??
$endgroup$
– MATH ASKER
Aug 26 '15 at 0:31












$begingroup$
Sure! I got the that the derivative of P(x) is 2ax+b through some rules of differentiation: the derivative of $ax^2$ is $2xa$, the derivative of $bx$ is $b$, and the derivative of $c$ is $0$ because $c$ is a constant. Thus the derivative of the whole polynomial is $2ax+b+0$. $P(2)=4$ was part of the information you gave, because you said that there is a maximum at $(2,4)$, which means that at $x=2$, we must have $y=4=P(2)$.
$endgroup$
– Aitor Ormazabal
Aug 26 '15 at 8:15




$begingroup$
Sure! I got the that the derivative of P(x) is 2ax+b through some rules of differentiation: the derivative of $ax^2$ is $2xa$, the derivative of $bx$ is $b$, and the derivative of $c$ is $0$ because $c$ is a constant. Thus the derivative of the whole polynomial is $2ax+b+0$. $P(2)=4$ was part of the information you gave, because you said that there is a maximum at $(2,4)$, which means that at $x=2$, we must have $y=4=P(2)$.
$endgroup$
– Aitor Ormazabal
Aug 26 '15 at 8:15


















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