A generic point on the graph of the curve has coordinates
[tex](x, 4x^2+1)[/tex]
The derivative gives us the slope of the tangent line at a given point:
[tex]f(x) = 4x^2+1 \implies f'(x) = 8x[/tex]
Let k be a generic x-coordinate. The tangent line to the curve at this point will pass through [tex](k, 4k^2+1)[/tex] and have slope [tex]8k[/tex]
So, we can write its equation using the point-slope formula: a line with slope m passing through [tex](x_0, y_0)[/tex] has equation
[tex]y-y_0 = m(x-x_0)[/tex]
In this case, [tex](x_0, y_0)=(k, 4k^2+1)[/tex] and [tex]m=8k[/tex], so the equation becomes
[tex]y-4k^2-1 = 8k(x-k)[/tex]
We can rewrite the equation as follows:
[tex]y-4k^2-1 = 8k(x-k) \iff y = 8kx - 8k^2+4k^2+1 \iff y = 8kx-4k^2+1[/tex]
We know that this function must give 0 when evaluated at x=0:
[tex]f(x) = 8kx-4k^2+1 \implies f(0) = -4k^2+1 = 8 \iff -4k^2 = 7 \iff k^2 = -\dfrac{7}{4}[/tex]
This equation has no real solution, so the problem looks impossible.
