A simple calculus question please

This line passes through the origin and touches the curve y=x^2+2 in the first quadrant. What is the equation of this line?


Please may I have the methods and explanations for the answer as well =)

This is a calculus question, so we're interested in rates of change, right? You want to know where the line y1 = kx + b touches the curve y2 = x^2 + 2, not intersects it. This means dy1/dx must equal dy2/dx if they're just touching (kissing). y1 = kx since b = 0 (passing through the origin). So:
dy1/dx = dy2/dx gives us k = 2x (take the derivatives of both y1 and y2 and set them equal. We also have the equation y1 = y2, which gives us kx = x^2 + 2. So now we have two equations and two variables, k and x to give us the touch point coordinates. k = 2x, so substitute this into kx = x^2 + 2 -->
2x^2 = x^2 + 2. Solve and get x = sqrt(2). This means the touch point coordinates are (sqrt(2), 4). substitute into y1 = kx and get k = 2 * sqrt(2), so y1 = 2*sqrt(2)x

i) Since the line passes through origin its equation is y = mx, where m is its slope.

ii) Solving this with the given y = x² + 2,

==> x² - mx + 2 = 0

Since the line y = mx is a tangent to y = x² + 2, the equation x² - mx + 2 = 0, should have an unique solution.
Thai is, its discriminant D = b² - 4ac = 0
==> m² - 8 = 0
So, m = ± √8
But given it touches the graph of the given function in 1st quadrant. The function y = x² + 2 being a parabola opening upwards, with its vertex at (0, 2), the slope of the tangent line is positive. Hence m = √8, that is m = 2√2

Thus the equation of the line is: y = (2√2)x.

This line touches the curve at (√2, 4).

y=3x

...you should really do the work yourself...

Practice makes perfect...

no one there to help you on the test/exem