point of intersection
Consider the function f(x) = -2 x^2 + 9 x + 11
(a) find [f(x+h) - f(x)]/h in the form __x+ __h+___
(I found -4x-2h+9, which was correct)
(b) Using the result of (a), find the derivative of f at x.
(I found -4x+9, also correct)
***(c) Find the point of intersection of the two tangent lines to the parabola y = -2x^2 + 9x + 11 at the points (3,20) and (-4,-57)
How do i do this? What are the two tangent lines they are talking about?
I don't know where to start on this bad boy, thanks so much
note that and are both on the curve
find the tangent line equation to the curve at each point, then set them equal to find where they intersect.
To find the tangent to a line you need to find the gradient of the tangent which is done by first differentiating the equation of the parabola:
To find the gradient at that point you need to put in the value of the x co-ordinate at the point of the tangent. In this case one of the values would be x = 3
Finally to find the equation of the tangent you now use the co-ordinates and the gradient you have just found and substitute those into the equation of a line:
This is the equation of the tangent at (3,20). To find the tangent at (-4, -57) you need to repeat this process using (-4,-57) instead of (3,20).
Arrange the equation of the second to equal y. You can then say both y values are the same so equate both tangents and solve to find the x value. Finally put this x value into either of the equations for the tangents and you will have the y value. This is your point of intersection.
I'll give you the chance to work out the second tangent yourself by following the same process I used. If you need me to explain further just ask.
how do you differentiate y= -2x^(2) + 9x+11?
I've done it above. See the part
Originally Posted by Amanda H
you already found the derivative, or didn't you notice?
Originally Posted by williamb
i have the point -1/2 ,30.5
which is correct
thank you both so much =]