Help With Finding Parabola

WhatIsMath

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Find parabola with equation y=a(x^2)+bx+c that has slope 8 at x=1, has slope -16 at x=-1, and passes through point (2,20)
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I am having issues with this problem in regards to finding the two points using the slopes. I understand how to solve the equation when having the 3 points, but can't figure out how to get the two points using the slope and x. Any help is greatly appreciated.
 

lev888

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What does it tell us about the function if we know the slope of its tangent at a particular point?
 

WhatIsMath

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I think that's where I'm confused as I don't know what the particular point is. I believe then I could use the slope to find the equation of the tangent line? I'm guessing the point would be (1,y) for the first one, but my confusion is around how to get that y for the tangent line.
 

lev888

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I think that's where I'm confused as I don't know what the particular point is. I believe then I could use the slope to find the equation of the tangent line? I'm guessing the point would be (1,y) for the first one, but my confusion is around how to get that y for the tangent line.
To find the points you only need to evaluate the function at the given x values.
What the slope gives you is the value of the derivative. So use the derivative to set up 2 equations.
 

WhatIsMath

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Could you please help me set up the formula for that? Sorry, I'm new to calculus as a college freshman so I want to make sure I go through the process correctly
 

tkhunny

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Why are you concerned with the location of points?

You have three parameters to find. You have three pieces of information. Use the information.

One for free: "passes through point (2,20)" ==> 20=4a+2b+c

For the next two, we're talking about the slope. This should trigger thoughts about the first derivative. Let's see what you get.
 

Jomo

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The derivative of f(x) is the slope formula for the function of f(x).
What does that mean? You have a function f(x) and you look at the graph of it. You pick out a certain point on the graph and notice that the x-value is say 1. Then you wonder what the slope of the tangent line at that point is. The answer is f'(1).
The slope of the tangent line of f(x) at x=a is f'(a). The slope of the tangent line of f(x) at x=a is f'(a). Th slope of the tangent line of f(x) at x=a is f'(a). Keep saying this until it sticks in your head. Of course you need to understand what it means as well.

f(x)=a(x^2)+bx+c that has slope 8 at x=1.
This means that f'(1) = 8. Can you write f'(1) in terms of a, b and c? If you set that expression equal to 8 then you will have 1 equation. Do the same with y=a(x^2)+bx+c has a slope of -16 at x=-1. This will give you a 2nd equation. You get the 3rd equation from knowing that the point (2,20) is on the curve f(x).
 

HallsofIvy

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I think that's where I'm confused as I don't know what the particular point is. I believe then I could use the slope to find the equation of the tangent line? I'm guessing the point would be (1,y) for the first one, but my confusion is around how to get that y for the tangent line.
You don't need to get the y value at that point! You are told that the curve has equation y= ax^2+ bx+ c so the slope at any given x is y'= 2ax+ b. Knowing that the curve "has slope 8 at x=1" tells you that y'(1)= 2a(1)+ b= a+ b= 8. Knowing that the curve "has slope -16 at x= -1" tells you that y'(-1)= 2a(-1)+ b= -2a+ b= -16. And, of course, knowing that the curve "passes through (2, 20) tells you that y(2)= a(2^2)+ b(2)+ c= 4a+ 2b+ c= 20.

Solve the three equations, a+ b= 8, -2a+ b= -16, and 4a+ 2b+ c= 20 for a, b, and c.
 

Li Batton

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First, take the derivative of

\(\displaystyle y=ax^2+bx+c\)

\(\displaystyle \frac{dy}{dx}=2ax+b\)

You are given two slopes

\(\displaystyle \frac{dx}{dx}=8\) when \(\displaystyle x=1\)

\(\displaystyle \frac{dy}{dx}=16\) when \(\displaystyle x=-1\)

Plugging these values into the derivative we will have two equations with two unkowns.

\(\displaystyle 8=2a(1)+b\)

\(\displaystyle 16=2a(-1)+b\)

solving this system will give you the values for a and b.

Then take the point (2,20) , a and b, plug them into

\(\displaystyle y=ax^2+bx+c\) and solve for c
 

WhatIsMath

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Thank you all for the tremendous help. What I was missing was y'(x)=slope. I have solved the problem correctly now.
 

Jomo

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Thank you all for the tremendous help. What I was missing was y'(x)=slope. I have solved the problem correctly now.
Yes, you are correct that you did not know that y'(x) gives the slopes. The problem is that this is key to calculus 1. You may be the best at finding derivatives but if you do not know what a derivative means then you really have not learned calculus 1 at all (not that you are done with calculus 1 yet). I am just trying to point out that if a student just learns one thing in calculus 1 it should be that the function f'(x) is the slope formula for f(x).
 
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