Question:

Step 1: Determine the signs of $\alpha$, $\beta$, and $\gamma$ from the graph. The parabola opens downward, which implies that the leading coefficient $\alpha$ is negative. $$ \alpha < 0 $$ The parabola intersects the $y$-axis above the origin. The $y$-intercept occurs at $x=0$, so $y = \alpha(0)^2 + \beta(0) + \gamma = \gamma$. Thus, $\gamma$ is positive. $$ \gamma > 0 $$ The axis of symmetry is given by $x = -\frac{\beta}{2\alpha}$. From the figure, the vertex of the parabola is in the second quadrant, meaning the axis of symmetry is to the left of the $y$-axis. $$ -\frac{\beta}{2\alpha} < 0 $$ Since $\alpha < 0$, it follows that $2\alpha < 0$. For the fraction $-\frac{\beta}{2\alpha}$ to be negative, and its denominator $2\alpha$ is negative, the numerator $-\beta$ must be positive. $$ -\beta > 0 \implies \beta < 0 $$ Thus, the signs are $\alpha < 0$, $\beta < 0$, and $\gamma > 0$. Step 2: Evaluate the given expressions. 1. **Expression involving $\alpha\beta$**: Since $\alpha < 0$ and $\beta < 0$, their product is positive. $$ \alpha\beta > 0 $$ 2. **Expression involving $\alpha^2 + \beta\gamma$**: Since $\alpha \neq 0$, $\alpha^2$ is positive. $$ \alpha^2 > 0 $$ Since $\beta < 0$ and $\gamma > 0$, their product $\beta\gamma$ is negative. $$ \beta\gamma < 0 $$ The expression $\alpha^2 + \beta\gamma$ is the sum of a positive number and a negative number. Its sign cannot be definitively determined without specific values for $\alpha$, $\beta$, and $\gamma$. For example, if $\alpha=-1, \beta=-1, \gamma=1$, then $\alpha^2+\beta\gamma = (-1)^2+(-1)(1) = 1-1=0$. If $\alpha=-2, \beta=-1, \gamma=1$, then $\alpha^2+\beta\gamma = (-2)^2+(-1)(1) = 4-1=3>0$. If $\alpha=-1, \beta=-2, \gamma=1$, then $\alpha^2+\beta\gamma = (-1)^2+(-2)(1) = 1-2=-1<0$. Therefore, this expression is not necessarily positive. 3. **Expression involving $\beta + \gamma - \alpha$**: Consider the value of the quadratic function at $x=1$, which is $f(1) = \alpha(1)^2 + \beta(1) + \gamma = \alpha + \beta + \gamma$. From the graph, at $x=1$, the parabola is above the $x$-axis, so $f(1) > 0$. The expression $\beta + \gamma - \alpha$ can be rewritten as $f(1) - 2\alpha$. $$ \beta + \gamma - \alpha = (\alpha + \beta + \gamma) - 2\alpha = f(1) - 2\alpha $$ We know $f(1) > 0$. Also, since $\alpha < 0$, it follows that $-2\alpha > 0$. Therefore, the sum of two positive numbers is positive. $$ \beta + \gamma - \alpha > 0 $$ 4. **Expression involving $\alpha\beta\gamma$**: We have $\alpha < 0$, $\beta < 0$, and $\gamma > 0$. The product of two negative numbers is positive, and the product of a positive number and a positive number is positive. $$ \alpha\beta\gamma = (\text{negative}) \times (\text{negative}) \times (\text{positive}) = (\text{positive}) \times (\text{positive}) = \text{positive} $$ $$ \alpha\beta\gamma > 0 $$