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Help on homework

cmodlin2

I have been stuck on this problem since Monday and cannot figure out what I am doing wrong. C

an someone help please.

abjorkma

I have the same problem but with different numbers. I can’t figure it out either!

ashley

For finding the y-intercept, I did the Qs y-coordinate plus the (slope times the y-intercept of the parabola). That will be your y-intercept number. It should be less than 29 but more than 12. Hope this helps. I am not the best at explanations.

mark

First off, I’d recommend that you use the point-slope formula for a line:

y-y_0=m(x-x_0).

Thus, if know the slope m and a point (x_0,y_0) that’s on the line, then you’ve got your formula.

Now, you can find two points on the line. The first (easier one) is when x=3 so that y = 3^2 + 3=12. Thus, (3,12) is on the line. The second is when y=29, yielding the equation

29 = x^2 + 3 \text{ or } x = \pm\sqrt{26}.

The picture seems to indicate that we should choose the negative root. Thus, our points are

(3,12) \text{ and } (-\sqrt{26},29)

and the slope is

m = \frac{29-12}{-\sqrt{26}-3}.

Using the first point (3,12) as the point on the line, I get

y-12 = \frac{29-12}{-\sqrt{26}-3}(x-3).

cmodlin2

I put that in and it still says incorrect

mark

Well, the correct way to type that in would be

12 + ((29-12)/(-sqrt(26)-3))*(x-3)

I recommend that you type it in full symbolic form like that. This particular problem would accepts a numeric approximation, which would be

-2.099*x+18.297

One final note, after looking over your WebWork input, is that WebWork doesn’t accept LaTeX input as we use here in the forum.

abjorkma

Hi, I’m having trouble with a and c from this problem, can someone help me out? Thanks!

mearing

That problem was hard for me to understand as well. All part “a” is asking for is for you to draw a line between (x_0,y_0) and (x_6,y_6) and then write the slope of that line \frac{f(x_2)-f(x_1)}{x_2-x_1}. It will look weird because the terms are x_0,y_0,x_6,y_6.

For part c. If you draw a tangent line line on (x_4,y_4), the line will almost touch the portion of curve directly below it and the slope will be pretty steep. I found it easier to picture (x_3,y_3) as the origin. So the other point would be (x_4,y_4) if it was reflected about the origin. Make sure to type the the answer like x=x_? rather than just x=?.