Solve the quadratic equation 8x^2=8x-16 to find the x coordinates of the intersection points.
you will find -1 and 8 . Substitute these values of x into the 16x to get the two values of the gradient.......
I am working through some introductory calculus. so I have a question (see attachment Q 56.)
. I got 16x from y=8x^2 as the gradient function. What I want to know is, am I supposed to determine the x coordinates from these two functions using the gradient function?
I just graphed the two functions and found their intercepts for the coordinates thus found the gradients for the two points this way.
I'd like help with Q 58 also please.
Firstly we know that the derivative gives us the gradient of the tangent to the curve at a particular point.
Let us consider your question, Q No 58.
equation of the curve is y = a x^4 Differentiating w r t x we get
y' = 4ax^3. This gives us slope of tangent to the curve. It is given that the gradient of the tangent at ( 3,b) is 2.
Gradient at ( 3,b) = 4 * a * 3 ^ 3 = 27*4 a = 2 given That gives 2 = 1/54
Also point ( 3, b ) lies on the curve and so must satisfy it
Thus we have from y = a * x^ 4
b = a * 3 ^ 4= 81 a OR b = 1/ 54 * 81 = 3/2. Hence we get a = 1/ 54 and b = 3/2