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For most Quadrátics in the form of ax^2+bx+c=0there is atleast 2 solutions in most cases as the graph of the quadratic passes the x axis on most cases. But In some cases the quadratic graph doesn’t Cross ...
The curve y = x^2 - 4 is a parabola that crosses the x axis at x = - 2 and x = 2, so the area that we are looking for is the area within the parabola when y <= 0 and -2<= x <= 2. So we expect ou...
Method 1: Quadratic formulaf(x)= ax2+bx+c, therefore a=1, b=12, c=32 - quadratic formula= (-b+/-(b2-4ac)1/2)/2ax=(-12+(144-128)1/2/2 = -4 or x=(-12-(144-128)Answered by • Maths tutor2876 Views
To find stationary points, we need to find dy/dx and set it equal to 0. Here we must use implicit differentiation: d/dx(x2) + d/dx(-xy) + d/dx(y2) = d/dx(12). Hence 2x - x(dy/dx) - y...
f(x) = x3 + 2x2 - x - 2 Use the factor theorem to test for f(1) = 0 f(1) = 13 + 2*12 - 1 - 2 = 0 Therefore x = 1 is a solution and (x-1) is a factor of f(x) Now...
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