Vtyjkyuk

$$y(x)= x^2-6x+22.09$$

I need to find the exact points where the tangents that pass through $0$, touch the curve using the first derivative. Any help would be great.

This is the parabola and the two tangent lines that pass through $(0,0)$:

In order to find the two tangent lines, let's consider the line equation in the slope-intercept form:
$$
y = ax + b
$$
Since both lines pass through $(0,0)$ we can easily deduct that $b=0$, so we have:
$$
y = ax
$$
In that equation, $a$ is the slope of the tangent line, which means it is its derivative:
$$
a = cfracdydx = 2x - 6
$$
So now we have a line equation that looks like this:
$$
y = ax = (2x - 6)x = 2x^2 - 6x
$$
Let's consider a point that passes through the parabola equation: $(x, x^2 - 6x + 22.09)$. Let's substitute this point in the tangent line equation:

$$
2x^2 - 6x = x^2 - 6x + 22.09 \
x^2 = 22.09 \
x = pm4.7
$$

Finally we have:

$y = 2x^2 - 6x rightarrow y = 2(4.7)^2 - 6(4.7) rightarrow y = 15.98$

$y = 2x^2 - 6x rightarrow y = 2(-4.7)^2 - 6(-4.7) rightarrow y = 72.38$

So the tangent lines intercept the parabola in the points $(4.7, 15.98)$ and $(-4.7, 72.38)$

Assuming I'm correct about what you are asking (see my comment), here is an outline of what you need to do. Pick some point $(x_0,y_0)$ on the curve an assume that the tangent to the curve at that point passes through $(0,0).$ Note that you are really only picking $x_0$ because once you've chosen $x_0$ that value of $y_0$ is determined. Now we knoe a point on the tangent line, namely $(x_0,y_0)$ and we know then slope, namely $y'(x_0).$

That gives you the equation of the tangent line, so you just need to see what condition ob $x_0$ will force $(0,0)$ to lie on the line. That's easy; just assume $(0,0)$ satisfies the equation, and see what that tells you about $x_0$. Check to be sure that the steps are reversible.

Let the equation of the tangent line be $y=mx$

The line and the parabola intersect where we have $$x^2-6x+22.09= mx$$double root.

That is $$x^2-(m+6)x+22.09=0$$ has double roots.

We need $$(m+6)^2-4(22.09)=0$$ which gives us$$ m=-6pm sqrt 22.09 =-6pm 4.7$$

For each of these slopes you find the intersection of $y=mx$ and your parabola.