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Find the x-coordinates of the stationary points on the graph with equation f(x)= x^3 + 3x^2 - 24x

This answer has 3 steps:1) find the derivative, f’(x), of the function2) factorise the derivative 3) set equal to zero and solve for x1) f’(x) = 3x^2 + 6x - 242)f’(x) = 3(x^2 + 2x - 8)3) x^2 + 2x - 8 = 0 (x+...
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Answered by Zaynab J. • Maths tutor
4335 Views

Determine for what values of c, f(x)=4x^2-(2c+8)x+4 has no real roots.

For f(x) not to have any real roots, its discriminant, b 2 -4ac < 0. Plugging in the coefficients from f(x), this means (-(2c+8)) 2 -4x4x4 < 0.This means (-(2c+8)) 2 < 4x4x4So 4c 2 +32c+64 < 64 b...
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Answered by Alexander R. • Maths tutor
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Show that the two vectors A= 2i+3j-k and B=3i-j+3k are perpendicular

We know that two vectors are perpendicular when their dot product is equal to 0. As you can see from the SQA Higher Maths formula sheet, the equation for the dot product is; A.B=a1b1+a2b2+a3b3. To answer thi...
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Answered by Samantha C. • Maths tutor
6877 Views

A circle has equation x^2+y^2-8x+10y+41=0. A point on the circle has coordinates (8,-3). Find the equation of the tangent to the circle passing through this point.

From the equation x 2 +y 2 -8x+10y+41=0, we can find the centre of the circle by the middle two terms -8x+10y and multiplying them by -1/2. So the coordinates of the circle centre are (4,-5).Now we can find ...
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Answered by Joseph K. • Maths tutor
2601 Views

Find an equation for the straight line AB , giving your answer in the form px+qy=r, where p, q and r are integers. Given that A has co-ordinates (-2,4) and B has co-ordinates (8,-6)

We know 2 equations for a straight line; Y=MX+C and Y-B=M(X-A) . Since we do not have the y intercept (C) or any means of finding it but we do have 2 points on the straight line, it would make sense to use t...
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Answered by Samantha C. • Maths tutor
4741 Views