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In general, the equation of a line is y-y0=m(x-x0), where m is the gradient and (x0,y0) is a point on the line. We need to find the gradient and a point on our tangent. Step 1: Find gradient. Since our ta...
The question asks for the roots, this simply means the numbers that could replace x in the equation and the result = 0. There are two ways too do this, firstly either using the Quadratic formula, or my fa...
(a) We have f(x)=(2/3)x3+ bx2 + 2x +3. First we have to solve for f'(x)=(dy/dx)f(x) which gives:f'(x) = [(32)/3]x2 + 2bx + 2 <=> f'(x)=2x2+ 2bx + 2 Fr...
To start, we need to complete the square of the equation. To do this, we divide the coefficient of x by 2. Here, 6/2=3. We then find (x+3)^2, which gives us the first part of the equation we want to expre...
When you find the first derivative of the equation and equate it to 0 then solve for x, you will find the x-coordinates of the stationary point(s) of the graph. For example, let's say f(x)=-x3+...
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