The correct answer is $6$ . It’s given that a line with equation $2 y equals 4 period 5$ intersects a parabola with equation $y = - 4 x^{2} + b x$ , where $b$ is a positive constant, at exactly one point in the xy-plane. It follows that the system of equations consisting of $2 y equals 4 period 5$ and $y = - 4 x^{2} + b x$ has exactly one solution. Dividing both sides of the equation of the line by $2$ yields $y equals 2 period 25$ . Substituting $2.25$ for $y$ in the equation of the parabola yields $2.25 = - 4 x^{2} + b x$ . Adding $4 x squared$ and subtracting $b x$ from both sides of this equation yields $4 x^{2} - b x + 2.25 = 0$ . A quadratic equation in the form of $a x^{2} + b x + c = 0$ , where $a$ , $b$ , and $c$ are constants, has exactly one solution when the discriminant, $b squared minus 4 a c$ , is equal to zero. Substituting $4$ for $a$ and $2.25$ for $c$ in the expression $b squared minus 4 a c$ and setting this expression equal to $0$ yields $b^{2} - 4 (4) (2.25) = 0$ , or $b^{2} - 36 = 0$ . Adding $36$ to each side of this equation yields $b squared equals 36$ . 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 $6$ .