Points of intersection...?
A is the point with coordinates (1,4) on the curve y=4x^2. B is the point with coordinates (0,1).
The line through A and B intersects the curve again at the point C. Show that the coordinates of C are (-1/4, 1/4).
Use calculus to find the equation of the tangent to the curve at A and verify that the equation of the tangent at C is y=-2x-1/4.
The two tangents intersect at the point D. Find the y-coordinate of D.
Stacey Warren - Expert brainly.com
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its a great way to start this problem by drawing it out and seeing what is happening. y = 4x^2 is a parabola with its vertex at (0,0), if you want roughly draw that out.
then draw the two points (1,4) and (0,1) are
you have two equations y = 4x^2 and another is y = rise *x / run + intercept or y = mx + b
long problem, first solve for the line between a and b. Then veirfy it is correct to the values of c. Then take the derivative of the curve. The derivative is the slope of the tangent line, and if you plug in x at say (1,4) you will get m=8.. To find the eq of the of the tangent line, use the point slope method and the line eq y=mx +b. at (1,4) the eq would look like 4=8(1)+b. once you solve for B then you have an equation of the tangent line at the point. You would just then repeat the steps at the points they asked of you, and verify the values that are necessary. If two lines intersect then their x and y values will be the same.
bringing it all together if you set both equations for the lines equal to each other you will get the points they are equal to each other or in other words where the cross soo
y = 4x^2
y = 3x + 1
==> 4x^2 = y = 3x +1
4x^2 = 3x + 1
4x^2 - 3x - 1 = 0
use the quadratic equation to solve
you know the line equeation can you also use y-y1=m(x-X1)
yeah you can use that also. the thing about this problem is the intercept is given by the point b because b is (0,1) understand?
yup, thanks :)
but that is a good way to go if b was another point other than the intercept :D
so now take the derivative of y = 4x^2, inorder to find the rate of change of y, then you can plug in any x value to find the slope at that point. the point of interest you are looking for is at A which has a x value of 4.
what do you get for the derivative
i dont get what a rise and run is
its kind of a cheap way of say the slope.
rise is the y value and run is the x value. its said "rise over run" it kind of flows well. you know how to get the slope from the two points A and B?
one more question
how did you come up with
y = 4x^2
y = 3x + 1?
well the equation given was y = 4x^2
the equation y = 3x + 1 is all gotten from the two points A and B
so you setup y = mx + b, find m = "rise over run" so its
m = \[\Delta y/\Delta x\]
the triangle means change in. or y2 -y1 => from (0,1) and (1,4)
take the y values so rise (or triangle y) = 4 - 1 = 3
y = 3x + b
now b we got from the point B (0,1) so b = 1
b is the y intercept
so y = 3x + 1
kinda get it?
i have to go soon not to rush you
its alright thanks ill ask my friends tomoz, coz its confusing :S
because i dont know why im putting it ina quadratic formula and what i am actually doin :(