The graph of the parabola $y = x^2 + 2bx + c$ is tangent to the line $y = 9x - 7$, where $c \neq 0$. What is $c-4b$?
This step is wrong:
4b^2 - 36b - 4c + 53 = 0
c = b^2 - 9b + 13
since 4*13 is not 53.
To find the value of $c-4b$, we need to determine the values of $b$ and $c$ first.
Since the parabola $y = x^2 + 2bx + c$ is tangent to the line $y=9x-7$, we know that the equation of the parabola can be written as $y = (x - a)^2 + k$, where $(a, k)$ is the point of tangency.
Setting the two equations equal to each other, we have:
$$(x - a)^2 + k = 9x - 7.$$
Expanding the left side, we get:
$$x^2 - 2ax + a^2 + k = 9x - 7.$$
Combining like terms, we have:
$$x^2 - (2a + 9)x + a^2 + k + 7 = 0.$$
Since the parabola is tangent to the line, the discriminant of this quadratic equation must be zero. Therefore:
$$\Delta = (-2a - 9)^2 - 4(a^2 + k + 7) = 0.$$
Expanding and simplifying, this becomes:
$$4a^2 + 36a + 81 - 4a^2 - 4k - 28 = 0.$$
Combining like terms again, we get:
$$32a + 53 - 4k = 0.$$
Isolating $k$, we have:
$$k = 8a + \frac{53}{4}.$$
We are given that $c \neq 0$, so $k \neq 0$. This means $8a \neq -\frac{53}{4}$.
Substituting this value of $k$ back into the equation of the parabola, we have:
$$y = (x - a)^2 + 8a + \frac{53}{4}.$$
The coefficient of $x$ in the equation of the parabola is $2b$. Therefore, we have:
$$2b = 8a \quad \Rightarrow \quad b = 4a.$$
Now we can substitute this value of $b$ back into the equation for $k$:
$$k = 8a + \frac{53}{4} = 2b + \frac{53}{4} = 2(4a) + \frac{53}{4} = 8a + \frac{53}{4}.$$
We obtained the same expression for $k$ as before, so we can use it to find $c-4b$.
Since $k = 8a + \frac{53}{4}$, we have:
$$c - 4b = (8a + \frac{53}{4}) - 4(4a) = (8a + \frac{53}{4}) - 16a = \frac{53}{4} - 8a.$$
Therefore, $c - 4b = \frac{53}{4} - 8a$.
To find the value of $c-4b$, we need to determine the values of $b$ and $c$. Since the graph of the parabola is tangent to the line $y=9x-7$, the equation of the line must be the same as the equation of the parabola at the point of tangency.
We have the equation of the line $y=9x-7$, and the equation of the parabola $y=x^2+2bx+c$. Substituting $y=9x-7$ into the equation of the parabola, we get:
$$9x-7 = x^2 + 2bx + c$$
Rearranging this equation, we have:
$$x^2 + (2b-9)x + (c+7) = 0$$
For the parabola to be tangent to the line, this quadratic equation must have a double root, which means its discriminant must be equal to zero.
The discriminant of this quadratic equation is given by the formula $b^2-4ac$. Since the equation must have a double root, we have:
$$(2b-9)^2 - 4(c+7) = 0$$
Expanding and simplifying, we get:
$$4b^2 -36b + 81 - 4c - 28 = 0$$
Simplifying further, we have:
$$4b^2 - 36b - 4c + 53 = 0$$
Now, we can equate the coefficients of the quadratic equation with zero to find the values of $b$ and $c$:
\begin{align*}
4b^2 - 36b - 4c + 53 &= 0 \\
4b^2 - 36b &= 4c - 53 \\
b^2 - 9b &= c - \frac{53}{4} \\
b^2 - 9b + \frac{81}{4} &= c - \frac{53}{4} + \frac{81}{4} \\
\left(b - \frac{9}{2}\right)^2 &= c + \frac{28}{4} \\
\left(b - \frac{9}{2}\right)^2 &= c + 7 \\
\end{align*}
From this equation, we can see that the value of $c-4b$ is equal to $7 - 4\left(\frac{9}{2}\right)$.
Simplifying this expression, we get:
$$c-4b = 7 - 18 = \boxed{-11}$$