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Let C : x^2-4x+2k be a parabola, with vertex m. By taking derivatives or otherwise discuss, as k varies, the coordinates of m and, accordingly, the number of solutions of the equation x^2-4x+2k=0. Illustrate your work with graphs

Write y=x²-4x+2k. And m:= (x_m, y_m) for the coordinates of our vertex. We deduce that x_m is exactly the value of x for which y'=2x-4=0, because m is a minimum p...

Answered by Massimiliano V. • Maths tutor

2584 Views

factorise x^3 + 3x^2 - 13x - 15

For convience we can call the polynomial f(x). Using the fact that the product of the roots of a polynomial equals the constant term, we know that the product of the roots is -15. We can therefore guess a...

Answered by Ben I. • Maths tutor

4912 Views

Sketch the curve y=x^2-x-6

This curve is a quadratic due to the highest power in the equation being two. Quadratics typically have the shape of a U. Due to the coefficient of the x^2 term being positive, the curve is increasing for...

Answered by Masum A. • Maths tutor

5497 Views

Find the intersection point of the line 2y=x+3 with the ellipse y^2+2x^2=3

The first step is to rearrange for x: we have x=2y-3now we can plug this into the equation of the ellipse: y^2+2(2y-3)^2=39y^2-24y+15 = 0we can use the quadratic formula to solve this equation:y = (24+-sq...

Answered by Amit B. • Maths tutor

2633 Views

Evaluate the integral between 5 and 3 for xe^x

Firstly, by looking at the integral, we see that this will have to be done by parts. There are 2 parts of this expression, the 'x' and the 'e^x'. As we know the formula f...

Answered by Ishaan O. • Maths tutor

2644 Views