In the x y -plane, a line with equation 2 y = 4.5 intersects a parabola

The correct answer is $6$. It’s given that a line with equation $2y=4.5$ intersects a parabola with equation $y=-4x^2+bx$, where $b$ is a positive constant, at exactly one point in the xy-plane. It follows that the system of equations consisting of $2y=4.5$ and $y=-4x^2+bx$ has exactly one solution. Dividing both sides of the equation of the line by $2$ yields $y=2.25$. Substituting $2.25$ for $y$ in the equation of the parabola yields $2.25=-4x^2+bx$. Adding $4x^2$ and subtracting $bx$ from both sides of this equation yields $4x^2-bx+2.25=0$. A quadratic equation in the form of $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants, has exactly one solution when the discriminant, $b^2-4ac$, is equal to zero. Substituting $4$ for $a$ and $2.25$ for $c$ in the expression $b^2-4ac$ and setting this expression equal to $0$ yields $b^2-4\left(4\right)(2.25)=0$, or $b^2-36=0$. Adding $36$ to each side of this equation yields $b^2=36$. Taking the square root of each side of this equation yields $b=\pm 6$. It’s given that $b$ is positive, so the value of $b$ is $6$.