Find b & c for y=3x²/4+bx+c with P(-1|9) & Tangent-slope=-6

[Elpinetos]

f: ℝ→ℝ y = 3x²/4+bx+c
Let P(-1|9) be a point on the function, with a tangent-slope of -6
Find b and c

No matter how I calculate, I get b = \frac{-9}{2} and c = \frac{15}{4}
The solution textbook I have tells me that c is \frac{-15}{4}

Am I doing something wrong or did they make a typo in the solution textbook? Oo

 

[Mark44]

Homework-type problems must be posted in the Homework & Coursework section, not in the technical math section. I am moving this post to the Calculus & Beyond section under Homework & Coursework.

 

[Mark44]

f: ℝ→ℝ y = 3x²/4+bx+c
Let P(-1|9) be a point on the function, with a tangent-slope of -6
Find b and c

No matter how I calculate, I get b = \frac{-9}{2} and c = \frac{15}{4}
The solution textbook I have tells me that c is \frac{-15}{4}

Am I doing something wrong or did they make a typo in the solution textbook? Oo

It would be helpful if you showed us what you did.

BTW, we write points with a comma separating the coordinates, not a vertical bar. In other words, like this: P(-1, 9).

 

[Elpinetos]

Okay, sorry :)

So what I did was, I got the first derivative y' = 3/2x+b

Then I set f(-1) = 9 and f'(-1) = -6

So -6 = 3/2*(-1)+b --> b = -9/2

9 = 3/4*(-1)² + (-1)*b + c

9 = 3/4 - b + c

9 = 3/4 - (-9/2) + c

9 = 3/4 + 18/4 + c

9-21/4 = c

c = 15/4

--> f: y = 3/4x^2 - 9/2x + 15/4

 

[Dick]

Okay, sorry :)

So what I did was, I got the first derivative y' = 3/2x+b

Then I set f(-1) = 9 and f'(-1) = -6

So -6 = 3/2*(-1)+b --> b = -9/2

9 = 3/4*(-1)² + (-1)*b + c

9 = 3/4 - b + c

9 = 3/4 - (-9/2) + c

9 = 3/4 + 18/4 + c

9-21/4 = c

c = 15/4

--> f: y = 3/4x^2 - 9/2x + 15/4

You are completely correct and there's a typo in the solution.

 

[Elpinetos]

Thank you :)

 

[Mark44]

It's easy enough to check. The values you got show that the function is f(x) = (3/4)x² - (9/2)x + 15/4. You can verify that f(-1) = 9 and that f'(-1) = -6.

Since your answer differs from the book answer, make sure that you are working the right problem and that you haven't overlooked or added a minus sign somewhere.