Basic maths slope question: find slope of f(X)=4x^2-5x+2 at (-1,11)

[Alexmcom]

Hey guys, I was just looking at an example in the book and I feel like I got my question incorrect
f(X)=4x^2-5x+2 at (-1,11) find the slope

f'(x)=8x-5

f'(X)=8(-1)-5=-8-5=-13

M=-13

m(x-x1)+y1
-13(x--1)+11
-13(x+1)+11
-13x-13+11
y=-13x-2

 

[ksdhart2]

Sorry, but I'm a bit confused as to what's going on here. If I'm understanding correctly, the problem asks you to find the slope of the given f(x) at the point (-1, 11). You took the derivative, which I agree is a correct step, given that the slope of a function at any point can be found be evaluating the derivative at that point. It turns out that, because the original function is a quadratic, that the derivative is itself a line. You found the slope of that line, but that's an unnecessary step. The slope of that line would be the second derivative of the original function. And then I'm not sure what the last steps you're showing are. The answer to the question, the slope of f(x) at the point x = -1, is given by evaluating f'(-1). Everything else you've done is unneeded.

 

[Alexmcom]

Sorry, but I'm a bit confused as to what's going on here. If I'm understanding correctly, the problem asks you to find the slope of the given f(x) at the point (-1, 11). You took the derivative, which I agree is a correct step, given that the slope of a function at any point can be found be evaluating the derivative at that point. It turns out that, because the original function is a quadratic, that the derivative is itself a line. You found the slope of that line, but that's an unnecessary step. The slope of that line would be the second derivative of the original function. And then I'm not sure what the last steps you're showing are. The answer to the question, the slope of f(x) at the point x = -1, is given by evaluating f'(-1). Everything else you've done is unneeded.

ahh I always fail to put in detail. My goal was to create a simple slope equation for the problem. The equation to the tangent line basically.

 

[ksdhart2]

Oh. Okay. Now I understand. Thanks for clarifying. Yes, the final equation you've given is the line that's tangent to f(x) at the point (-1, 11). You can verify this in one of two ways: By graphing both, and you'll see that they intersect at exactly one point: (-1, 11). Or you can do it algebraically, by setting the two expressions equal: -13x - 2 = 4x² - 5x + 2. You'll end up with a double root at x = -1, which also verifies that they intersect exactly once, at that point. Was there something in particular that made you doubt your answer?

 

[Alexmcom]

Oh. Okay. Now I understand. Thanks for clarifying. Yes, the final equation you've given is the line that's tangent to f(x) at the point (-1, 11). You can verify this in one of two ways: By graphing both, and you'll see that they intersect at exactly one point: (-1, 11). Or you can do it algebraically, by setting the two expressions equal: -13x - 2 = 4x² - 5x + 2. You'll end up with a double root at x = -1, which also verifies that they intersect exactly once, at that point. Was there something in particular that made you doubt your answer?

Si there was. I saw some notes and I tried to calculate it myself and I didn't get his answer at all so I was confused. Whenever I get the answer wrong I doubt my answer is correct so I try figure it out until I the answer is the same. Also, may I ask you another question? It's related to scalar, vector and paramedic equations. It's also a confirmation.

 

[ksdhart2]

Si there was. I saw some notes and I tried to calculate it myself and I didn't get his answer at all so I was confused. Whenever I get the answer wrong I doubt my answer is correct so I try figure it out until I the answer is the same. Also, may I ask you another question? It's related to scalar, vector and paramedic equations. It's also a confirmation.

Sure. I don't mind. I might not know the answer, but I'm certainly willing to give it a go.

 

[Steven G]

Hey guys, I was just looking at an example in the book and I feel like I got my question incorrect
f(X)=4x^2-5x+2 at (-1,11) find the slope

f'(x)=8x-5

f'(X)=8(-1)-5=-8-5=-13

My only concern is that you wrote that f'(x)=8x-5 and that f'(X)=8(-1)-5=-8-5=-13. Well which is it? Does f'(x)=8x-5 or does f'(X)=13?

After all f'(x) can't equal both those values-or can it.

What you meant to write was f'(-1)=8(-1)-5=-8-5=-13.

As far as I'm concerned, if the equal sign are not valid, then what you wrote is wrong.

 

[Otis]

you wrote that f'(x)=8x-5 and that f'(X)=8(-1)-5=-8-5=-13. Well which is it?

X is not the same symbol as x.

What was written shows that X represents the constant -1, where x is the variable.

By the way, one of your -13s is missing its negative sign.