Analyse the curve y = -2x^2 e^x: find its intercepts, turning points, and behaviour as x tends to plus and minus infinity.

Step-by-step solution

1. 1

   Establish the sign of the whole function first. For every real $x$ we have $x^2\ge0$ and $e^x>0$, so

   $-2x^2e^x\le0$

   The curve therefore lies entirely on or below the $x$-axis, touching it only where $x^2=0$. Knowing this before differentiating rules out half the sketch.

2. 2

   Find the intercepts. Setting $y=0$: since $e^x$ is never zero, we need $x^2=0$, so $x=0$. Both the $x$-intercept and the $y$-intercept are the single point $(0,0)$.

3. 3

   Differentiate with the product rule. With $u=-2x^2$ and $v=e^x$:

   $y'=-4xe^x+\left(-2x^2\right)e^x=-2e^x\left(x^2+2x\right)=-2e^xx(x+2)$

   Factoring out $-2e^x$ makes the critical points readable at a glance.

4. 4

   Locate the critical points. Since $e^x\neq0$, $y'=0$ requires $x(x+2)=0$:

   $x=0\qquad\text{or}\qquad x=-2$

   At $x=0$, $y=0$. At $x=-2$:

   $y=-2(-2)^2e^{-2}=-8e^{-2}=-\frac{8}{e^2}\approx-1.083$

5. 5

   Classify them from the sign of y'. The factor $-2e^x$ is always negative, so $y'$ has the opposite sign to $x(x+2)$: negative for $x<-2$, positive for $-2<x<0$, negative for $x>0$. So the curve falls, then rises, then falls — giving a local minimum at $\left(-2,-\tfrac{8}{e^2}\right)$ and a local maximum at $(0,0)$. The maximum value $0$ is also the global maximum, consistent with step 1.

6. 6

   Determine the end behaviour. As $x\to-\infty$, $e^x\to0$ far faster than $x^2\to\infty$ grows, so $y\to0^-$: the $x$-axis is a horizontal asymptote approached from below. As $x\to+\infty$, both factors grow and $y\to-\infty$, plunging steeply. So the shape is: hugging the axis from below on the far left, dipping to $-1.083$ at $x=-2$, back up to touch the origin, then falling away without bound.

7. 7

   Verify the key numbers. Symmetric difference quotients of step $10^{-7}$ match $y'=-2e^xx(x+2)$ at $x=-3,-1,0.5,2$ ✓. Sample values: $y(-4)=-0.586$, $y(-3)=-0.896$, $y(-2)=-1.083$, $y(-1)=-0.736$, $y(1)=-5.437$, $y(2)=-59.112$ — all recomputed and confirming both the dip at $x=-2$ and the steep right-hand fall ✓.