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Find the two points of intersection of the graphs 2x+y=7 and x^2-8x+7=y. Solve using only algebraic methods (no graphical).

We treat this problem as two simultaneous equations, using our knowledge that when two graphs intersect, they simultaneously have the same solution. The second equation (quadratic equation) is already in the...
AS
Answered by Amy S. • Maths tutor
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A curve C has equation y = x^2 − 2x − 24sqrt x, x > 0. Prove that it has a stationary point at x=4.

A stationary point is where the curve has 0 gradient. So to prove that x=4 is a stationary point, we must find the equation of the first derivative. To do this, differentiate x 2 - 2x - 24sqrtx. It might hel...
EF
Answered by Elizabeth F. • Maths tutor
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Find the second derivate d^2y/dx^2 when y = x^6 + sqrt(x).

Initially we find the first derivative of the function y = x 6 + sqrt(x). We achieve this by multiplying each x term by the power it is raised to, then reducing the power by 1 . Solution: 1) It helps to init...
Answered by • Maths tutor
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Solve 4(3x - 2) = 2x - 5. (3 marks)

To solve this linear equation, our ultimate goal is to end up with our unknown "x" value on one side and our number on the other side of this equation. To get to this stage, there are three steps.F...
HE
Answered by Hugh E. • Maths tutor
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The point P has coordinates (3, 4) The point Q has coordinates (a, b) A line perpendicular to PQ is given by the equation 3x + 2y = 7 Find an expression for b in terms of a

Perpendicular gradients multiply to give -1. The gradient of the perpendicular line (y=-3/2x+7/2) is -3/2 , so the gradient of PQ is 2/3. Using gradient formula change in y/ change in x :(4-b)/(3-a) = 2/3 Si...
LD
Answered by Lizzie D. • Maths tutor
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