We are given the quadratic curve $y = ax^2 + bx + c$, which passes through the point $(1,7)$ and is tangent to the line $y = 6x$ at the origin.

Let's break down the conditions:

1. Point (1,7) on the curve: When $x = 1$, $y = 7$, so substituting this into the quadratic equation gives: $a(1)^2 + b(1) + c = 7$ Simplifying: $a + b + c = 7 \quad \text{(Equation 1)}$

2. Tangent to the line $y = 6x$ at the origin:

   • At the origin $(0,0)$, both the curve and the line have the same value of $y$, so: $a(0)^2 + b(0) + c = 0 \implies c = 0$
   • Since the curve is tangent to the line at the origin, their slopes at the origin must also be the same. The slope of the line $y = 6x$ is 6. The derivative of the quadratic curve is: $\frac{dy}{dx} = 2ax + b$ At $x = 0$, the slope of the curve is $b$. Since the slope of the line is 6 at the origin, we have: $b = 6 \quad \text{(Equation 2)}$

3. Substitute $c = 0$ and $b = 6$ into Equation 1: $a + 6 + 0 = 7$ Simplifying: $a = 1 \quad \text{(Equation 3)}$

Thus, the values of $a$, $b$, and $c$ are: $a = 1, \quad b = 6, \quad c = 0$

Would you like any more details on how we solved the system of equations?

Here are some related questions for further exploration:

1. How do we derive the equation of a tangent line for any quadratic function?
2. What role does the derivative play in determining tangency between curves and lines?
3. Can you explain why the condition at the origin required both the function value and the slope to match?
4. How would the solution change if the curve passed through a different point instead of $(1,7)$?
5. What happens to the quadratic curve if the coefficients $a$, $b$, or $c$ take on different signs?

Tip: Always verify both the value and slope conditions when working with tangents between a curve and a line!