The point (0,1) lies in the "interior" of the parabola
y = 3x^2.
There is no tangent to the curve that will pass through (0,1).
check your typing.
y = 3x^2.
There is no tangent to the curve that will pass through (0,1).
check your typing.
let the point of contact be (a,b)
slope of tangent by the grade 9 way = (b+1)/a
slope of tangent by Calculus is
dy/dx = 6x
so at the point (a,b), slope = 6a
then 6a = (b+1)/a
6a^2 = b+1
but since (a,b) lies on the curve, b = 3a^2
so
6a^2 = 3a^2 + 1
a^2 = 1/3
a = ± 1/√3
so the x coordinates of the two tangents are 1/√3 and -1/√3
2x^2 + 4y^2 = 36
that passes through the points: 14,3
Step 1: Find the derivative of the function f(x) = 3x^2.
The derivative of f(x) gives us the slope of the tangent line at any given point on the curve.
In this case, f'(x) = 6x, as the derivative of 3x^2 with respect to x is 6x.
Step 2: Find the slope of the tangent lines at point P(0,1).
The slope of the tangent line passing through point P is given by the value of f'(x) at x = 0.
Substituting x = 0 in f'(x) = 6x, we get f'(0) = 6(0) = 0.
Step 3: Write the equation of the tangent line passing through P(0,1) using the slope-point form.
The general equation of a line passing through point (x1, y1) with slope m is given by y - y1 = m(x - x1).
Substituting the values y1 = 1, m = 0, and x1 = 0 into the equation, we get y - 1 = 0(x - 0), which simplifies to y - 1 = 0.
Step 4: Solve the equation of the tangent line with f(x) = 3x^2 to find the x-coordinates of the intersection points.
Substituting y = 3x^2 into the equation y - 1 = 0, we get 3x^2 - 1 = 0.
Rearrange the equation to isolate x: 3x^2 = 1.
Divide by 3: x^2 = 1/3.
Take the square root: x = ±√(1/3).
Therefore, the x-coordinates of the points where the tangent lines intersect the curve f(x) = 3x^2 are x = √(1/3) and x = -√(1/3).