Question 5314: I need to know the steps to solve this problem. Hope you can help me.
If f(x) = 3-x^2
how do I find the slope of the tangent line to the graph of f at (1,2)
How do I find the slope-intercept equation of the tangent line to the graph of f at (1,2)
How do I graph f and the tangent line.
Thank you
Sue
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! gradient is found by differentiation.
f(x) = 3-x^2 so f'(x) = -2x --> this is the gradient at any point on the curve. You need to find the gradient at x=1, so then... f'(x) = -2(1) which is -2.
Now the tangent at a point (by definition) has the same gradient as the curve at that point, so the general form of the straight line is y=mx+c. We know m=-2, so we have y=-2x+c. We need to find c. To do that we need to know the x and y, which we do! (1,2)...so 2 = -2(1) + c --> c=4. So the tangent has equation y=-2x+4
When you say graph, you mean sketch it? or do it on graph paper?
Sketch --> need to know what shape the quadratic is..its x^2 term is -ve so that means it is a n-shape hump. Where does it cross the x-axis and the y-axis? You find out.
For the straight line... at the point on the curve where x=1, draw the tangent. the line should cross the y-axis at y=4. Remember this is a sketch, so not perfect artistic skills required, but at least try to get the general jist of the shapes and layout :-)
jon.
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