<s>[INST] The line is tangent to a parabola The line $y = 4x-7$ is tangent to a parabola that has a $y$-intercept of $-3$ and the line $x=\frac{1}{2}$ as its axis of symmetry. Find the equation of the parabola. I really need help solving this question. THx
 [/INST] Let the formula for the parabola be $f(x)=ax^2+bx+c$
1. y-intercept: $y$-intercept is @ $(0,-3)$ so we have: $f(0)=-3 \Rightarrow c=-3$.
2. Vertex: The vertex is @ $(x,\frac{1}{2})$ and we now that at the vertex, the derivative is equal to zero, so we have:
$f'(\frac{1}{2})=0 \Leftrightarrow 2a\cdot(\frac{1}{2})+b=0 \Rightarrow a+b=0$.
3. Line tangent to parabola: Furthermore, the line $y=4x-7$ is tangent to the parabola (and therefore $4x-7=ax^2+bx+c$ has only one solution).
Intersection point: $ax^2+bx+c=4x-7 \Rightarrow ax^2+x(b-4)+(c+7)$. Discriminant must be zero in order to have only one intersection point, so we have $D=b^2-4ac=(b-4)^2-4a(c+7)=0$.
Now, we substitute $c=-3$ and $a=-b$ and solve for $a$:
$(b-4)^2+4b(-3+7)=0 \Rightarrow (b-4)^2+16b=0 \Rightarrow b=-4$. $a=-b$ so $a=4$.
The formula is $\boxed{y=4x^2-4x-3}$.

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