there are two tangents lines to the curve f(x) = 3x^2 that pass through the point p =0,1 find the x coordinates of the point where the tangents line intersect the curve

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.

there are two tangents lines to the curve f(x) = 3x^2 that pass through the point p =0,-1 find the x coordinates of the point where the tangents line intersect the curve, please show working.

ahhh, now it makes sense.

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

find the equation of the tangent line of the curve y=root x divideed by x+1 at the point p = 2, root 2 divided by 3. show working please.

find the equation of the tangent line to the curve y = 2√2/x+1 at the point p=1,2/3.please show working. i would appreciate if you do it before 12:05pm today

use implcit differenciation to find an eqaution of both the tangent line to the ellipse:

2x^2 + 4y^2 = 36
that passes through the points: 14,3

To find the x-coordinates where the tangent lines intersect the curve f(x) = 3x^2, we need to determine the equations of the tangent lines first. Let's proceed step by step:

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).