Integration- Area under Graph help? y = 4 - e^{(1/2)x} & tangent at axis

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claudiachm

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Hi, I'm new here. ^-^ I need a lot of help with calculus and I was wondering if somebody could help me with this question? I've subbed in y=0 into the equation but can't seem to get a valid answer. Thank you in advance!!
 
You may find it simpler by using the y-axis and x = f(y).
 
claudiachm said:
Hi, I'm new here. ^-^ I need a lot of help with calculus and I was wondering if somebody could help me with this question?


The curve \(\displaystyle y\, =\, 4\, -\, e^{\frac{1}{2}\, x}\) meets the y- and x-axes at points A and B, respectively.

(a) Find the coordinates of A and B.

(b) The tangent to the curve at point A meets the x-axis at point C. Find the coordinates of C.



I've subbed in y=0 into the equation but can't seem to get a valid answer.
Click to expand...
Part (a) requires only what you learned back in algebra.

To find the x-intercept, plug zero in for y, and solve the resulting exponential equation (here). It sounds as though you've done the substitution part. Where are you getting stuck on the solving part? Please show all of your steps.

To find the y-intercept, plug zero in for x, and simplify to find the coordinate.

Part (b) requires that you use some calculus. What does the keyword "tangent" suggest that you need to do with the given equation? Once you've done this, how should the result be combined with the point A? Now that you have a point and a slope, return to basic algebra to find the line equation (here). At point C, what must be the value of y? Where does this lead?

Please be complete. When you reply, please provide the rest of the information for this exercise (that is, the portion(s) which you believe require integration or the area under the curve, per your subject line). Thank you! ;)
 
stapel said:
Part (a) requires only what you learned back in algebra.

To find the x-intercept, plug zero in for y, and solve the resulting exponential equation (here). It sounds as though you've done the substitution part. Where are you getting stuck on the solving part? Please show all of your steps.

To find the y-intercept, plug zero in for x, and simplify to find the coordinate.

Part (b) requires that you use some calculus. What does the keyword "tangent" suggest that you need to do with the given equation? Once you've done this, how should the result be combined with the point A? Now that you have a point and a slope, return to basic algebra to find the line equation (here). At point C, what must be the value of y? Where does this lead?

Please be complete. When you reply, please provide the rest of the information for this exercise (that is, the portion(s) which you believe require integration or the area under the curve, per your subject line). Thank you! ;)
Click to expand...


Thank you so much!! I think I should find the gradient first, shouldn't I? ^-^
 
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