At what point is the slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ maximum? Also find the maximum slope.

> [!TLDR] > The slope of the curve is given by the derivative $m(x) = -3x^2 + 6x + 9$. This is a concave-down quadratic, so its maximum occurs at the vertex $x = 1$, and the maximum slope is $m(1) = 12$. The question asks: at what point on the curve is the slope itself the largest? That means we first need the slope function (the derivative), and then we need to maximise that function. **Why the Mean Value Theorem isn't the main tool here** — you might think of MVT because it talks about average slope, but this problem is about the *maximum* of a slope function. That's a pure optimisation problem: find where the derivative of the slope (the second derivative) is zero and check concavity. --- 1. **Find the slope function.** The slope of $y = -x^3 + 3x^2 + 9x - 27$ at any $x$ is $$m(x) = y' = -3x^2 + 6x + 9.$$ This is a quadratic in $x$, opening downward (coefficient of $x^2$ is negative). So its graph is an upside-down parabola — it has a single maximum point. 2. **Maximise the slope function.** For a quadratic $ax^2 + bx + c$ with $a < 0$, the maximum occurs at $$x = -\frac{b}{2a}.$$ Here $a = -3$, $b = 6$, so $$x = -\frac{6}{2(-3)} = -\frac{6}{-6} = 1.$$ That's the $x$-coordinate where the slope is greatest. Alternatively, set $m'(x) = 0$: $$m'(x) = -6x + 6 = 0 \implies x = 1.$$ Since $m''(x) = -6 < 0$, this is indeed a maximum. 3. **Find the maximum slope.** Plug $x = 1$ into $m(x)$: $$m(1) = -3(1)^2 + 6(1) + 9 = -3 + 6 + 9 = 12.$$ 4. **Find the point on the curve.** The question asks "at what point" — that means the coordinates $(x, y)$. We have $x = 1$. Find $y$: …

At what point is the slope of the curve y = -x³ + – Class 12 Mathematics

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y = -x^3 + 3x^2 + 9x - 27

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Concept understanding — Quadratic Extrema

Quadratic Extrema: From Intuition to Precision

Toss a ball straight up: it rises, slows, stops for an instant at the top, then falls. Plot its height against time and you get a parabola with exactly one turning point — a peak (maximum) or a valley (minimum). That single highest or lowest point is what quadratic extrema are about.

The Intuition First

A quadratic is $f(x) = ax^2 + bx + c$ , with $a \neq 0$ ; its graph is a parabola.

If $a > 0$ , it opens upward (a U) and has a minimum at the bottom.
If $a < 0$ , it opens downward and has a maximum at the top.

The turning point is the vertex. Every quadratic has exactly one vertex — that's the extremum.

Note

Unlike cubic or higher-degree polynomials, a quadratic never has both a maximum and a minimum. It has one or the other.

The Precise Statement

For $f(x) = ax^2 + bx + c$ with $a \neq 0$ :

Vertex (extremum) at

x = -\frac{b}{2a}

Extremum value

f\left(-\frac{b}{2a}\right) = c - \frac{b^2}{4a}

Nature: $a > 0$ → minimum; $a < 0$ → maximum.

\text{Vertex} = \left( -\frac{b}{2a},\; c - \frac{b^2}{4a} \right)

Why That $x$ ? A Quick Derivation

Complete the square:

f(x) = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right)

The squared term is always $\geq 0$ . When $a > 0$ , $f(x)$ is smallest when the square is zero — at $x = -\frac{b}{2a}$ . When $a < 0$ , the largest value occurs at the same $x$ .

Tip

The vertex's $x$ -coordinate is also the average of the two roots (if they exist): $x = \frac{\text{root}_1 + \text{root}_2}{2}$ .

Common Mistake to Avoid

Watch out

Don't confuse the sign of $a$ with the sign of the extremum value. With $a > 0$ you always have a minimum, but that minimum could be positive, negative, or zero. The shape tells you max vs min, not the number itself.

Example

Find the extremum of $f(x) = 2x^2 - 8x + 5$ . …

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The slope of the curve is given by the derivative $m(x) = -3x^2 + 6x + 9$ . This is a concave-down quadratic, so its maximum occurs at the vertex $x = 1$ , and the maximum slope is $m(1) = 12$ .

The question asks: at what point on the curve is the slope itself the largest? That means we first need the slope function (the derivative), and then we need to maximise that function.

Why the Mean Value Theorem isn't the main tool here — you might think of MVT because it talks about average slope, but this problem is about the maximum of a slope function. That's a pure optimisation problem: find where the derivative of the slope (the second derivative) is zero and check concavity.

Find the slope function. The slope of $y = -x^3 + 3x^2 + 9x - 27$ at any $x$ is

m(x) = y' = -3x^2 + 6x + 9.

This is a quadratic in $x$ , opening downward (coefficient of $x^2$ is negative). So its graph is an upside-down parabola — it has a single maximum point.

Maximise the slope function. For a quadratic $ax^2 + bx + c$ with $a < 0$ , the maximum occurs at

x = -\frac{b}{2a}.

Here $a = -3$ , $b = 6$ , so

x = -\frac{6}{2(-3)} = -\frac{6}{-6} = 1.

That's the $x$ -coordinate where the slope is greatest.

Alternatively, set $m'(x) = 0$ :

m'(x) = -6x + 6 = 0 \implies x = 1.

Since $m''(x) = -6 < 0$ , this is indeed a maximum.

Find the maximum slope. Plug $x = 1$ into $m(x)$ :

m(1) = -3(1)^2 + 6(1) + 9 = -3 + 6 + 9 = 12.

Find the point on the curve. The question asks "at what point" — that means the coordinates $(x, y)$ . We have $x = 1$ . Find $y$ : …

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Q.At what point is the slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ maximum? Also find the maximum slope.

Concept understanding — Quadratic Extrema

Quadratic Extrema: From Intuition to Precision

The Intuition First

The Precise Statement

Why That $x$ ? A Quick Derivation

Common Mistake to Avoid

Example

Method: Optimizing the Slope Function of a Curve

Steps

Common Mistakes

Q.At what point is the slope of the curve y=−x3+3x2+9x−27y = -x^3 + 3x^2 + 9x - 27y=−x3+3x2+9x−27 maximum? Also find the maximum slope.

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Q.At what point is the slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ maximum? Also find the maximum slope.

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