Question

which equation could generate the curve in the graph below?

y
x

y = -2x² + 3x - 5
y = -2x² - 4x - 2
y = -2x² - 16x - 28
y = -2x² + 6x - 28

Explanation:

Step1: Analyze parabola direction and vertex

The parabola opens downward (since the coefficient of \(x^2\) is negative for all options). The vertex is in the second quadrant (left of y - axis, above x - axis). Let's use the vertex formula \(x =-\frac{b}{2a}\) for a quadratic \(y = ax^{2}+bx + c\).

Step2: Analyze each option

• Option 1: \(y=-2x^{2}+3x - 5\), \(a=-2\), \(b = 3\). \(x=-\frac{3}{2\times(-2)}=\frac{3}{4}\) (positive x - coordinate, vertex on right of y - axis, eliminate).
• Option 2: \(y=-2x^{2}-4x - 2\), \(a=-2\), \(b=-4\). \(x =-\frac{-4}{2\times(-2)}=\frac{4}{-4}=-1\). Then \(y=-2(-1)^{2}-4(-1)-2=-2 + 4-2 = 0\). Vertex at \((-1,0)\), but the graph has vertex above x - axis, eliminate.
• Option 3: \(y=-2x^{2}-16x - 28\), \(a=-2\), \(b=-16\). \(x=-\frac{-16}{2\times(-2)}=\frac{16}{-4}=-4\). \(y=-2(-4)^{2}-16(-4)-28=-32 + 64-28 = 4\). Vertex at \((-4,4)\) (second quadrant, above x - axis).
• Option 4: \(y=-2x^{2}+6x - 28\), \(a=-2\), \(b = 6\). \(x=-\frac{6}{2\times(-2)}=\frac{6}{4}=\frac{3}{2}\) (positive x - coordinate, vertex on right of y - axis, eliminate).

Also, check x - intercepts. For option 3, set \(y = 0\): \(-2x^{2}-16x - 28 = 0\) divide by - 2: \(x^{2}+8x + 14 = 0\)? Wait, no, wait: \(-2x^{2}-16x - 28=0\) => \(x^{2}+8x + 14 = 0\)? Wait, no, let's factor: \(-2(x^{2}+8x + 14)=0\)? Wait, no, \(x^{2}+8x + 14\) discriminant \(64 - 56 = 8>0\), but wait, earlier calculation for vertex was \((-4,4)\). Wait, maybe I made a mistake in factoring. Wait, \(-2x^{2}-16x - 28=-2(x^{2}+8x + 14)\)? No, \(x^{2}+8x + 14\) is wrong. Wait, \(-2x^{2}-16x - 28 = 0\) => \(x^{2}+8x + 14 = 0\)? No, 28 divided by 2 is 14? Wait, no, \(-2x^{2}-16x - 28 = 0\) divide both sides by - 2: \(x^{2}+8x + 14 = 0\)? Wait, discriminant \(D = 64-56 = 8\), roots \(x=\frac{-8\pm\sqrt{8}}{2}=-4\pm\sqrt{2}\approx - 4\pm1.414\), so roots at \(-2.586\) and \(-5.414\), both negative (left of y - axis), which matches the graph.

For option 2, \(y=-2x^{2}-4x - 2=-2(x^{2}+2x + 1)=-2(x + 1)^{2}\), vertex at \((-1,0)\), which is on the x - axis, but the graph has vertex above x - axis, so option 3 is correct.

Answer:

\(y = -2x^{2}-16x - 28\) (the third option)