We look for the $y$-intercept of the graph, i.e., the value of $y$ when $x = 0$: the parabola passes through the point $(0,5)$, which is marked in red below. The equation $y = ax^2 + bx + 5$ then fits the red parabola. [asy]
defaultpen(linewidth(0.7));
draw((-1,0)--(6,0),Arrow);
draw((0,-1)--(0,11),Arrow);
for (int i = 1; i <= 5; ++i) {
draw((i,-0.2)--(i,0.2));
label("$"+string(i)+"$",(i,-0.2),S);
draw((-0.2,i)--(0.2,i));
label("$"+string(i)+"$",(0.2,i),E);
}
draw((0,5) parabola bend (2.5,8.75) (5,5));
dot("$(0,5)$",(0,5),W,linewidth(4));
[/asy] By inspection, we can see that the roots of the quadratic are $x = 1$ and $x = 4$, so if the quadratic factorizes as \[y = a(x - 1)(x - 4),\] then $a = 5$, by the value $(0,5)$ on the graph.
The coefficient of $x^2$ is $a = 5$. The constant term is obtained by substituting $(x,y) = (1,0)$, which gives $0 = a(1 - 1)(1 - 4) = -15a$, so $a = 5$ indeed. Thus, $a \cdot b \cdot c = 5 \cdot 1 \cdot -15 = \boxed{-75}$.
The parabola $y = ax^2 + bx + c$ is graphed below. Find $a \cdot b \cdot c$. (The grid lines are one unit apart.)
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1 answer