Question

which equation could generate the curve in the graph below?
options:

• $y = -2x^2 + 3x - 5$
• $y = -2x^2 - 4x - 2$
• $y = -2x^2 - 16x - 28$
• $y = -2x^2 + 6x - 28$

Explanation:

Step1: Analyze the parabola's direction and roots

The parabola opens downward, so the coefficient of \(x^2\) is negative (all options have \(-2\), so that's good). Now, check the roots (where \(y = 0\)) and the vertex. Let's analyze each option.

Step2: Analyze Option A (\(y=-2x^2 + 3x - 5\))

Find discriminant: \(D = 3^2 - 4(-2)(-5)=9 - 40=-31 < 0\). No real roots, but the graph has two real roots (crosses x - axis twice). Eliminate A.

Step3: Analyze Option B (\(y=-2x^2 - 4x - 2\))

Factor: \(y=-2(x^2 + 2x + 1)=-2(x + 1)^2\). This is a perfect square, so it touches the x - axis at \(x=-1\) (one root, multiplicity 2). But the graph has two distinct roots. Eliminate B.

Step4: Analyze Option C (\(y=-2x^2 - 16x - 28\))

Factor out \(-2\): \(y=-2(x^2 + 8x + 14)\). Find roots using quadratic formula: \(x=\frac{-8\pm\sqrt{64 - 56}}{2}=\frac{-8\pm\sqrt{8}}{2}=\frac{-8\pm2\sqrt{2}}{2}=-4\pm\sqrt{2}\approx - 4\pm1.414\), so roots are \(\approx - 2.586\) and \(\approx - 5.414\), both negative (matches the graph's x - intercepts on the left side of y - axis).

Step5: Analyze Option D (\(y=-2x^2 + 6x - 28\))

Discriminant: \(D = 6^2 - 4(-2)(-28)=36 - 224=-188 < 0\). No real roots. Eliminate D.

Answer:

\(y = -2x^2 - 16x - 28\) (Option C)