Recall that the gradient of a straight line (a tangent) is constant at any point on the line. However, the gradient of a curve changes from point to point along the curve.

Consider the curve of y = x^{2} + 2x – 4 above:

Note that the gradient of the points P and S is the same as the gradient of PQ and ST gradient, respectively.Â

Gradient at P = \( \frac{y_2 \: – \: y_1}{x_2 \: – \: x_1} \\ = \frac{3\: – \: (-1)}{2\: – \: 1} \\ = \frac {4}{1} \\ = \scriptsize 4\)

Also Gradient at S = \( \frac{y_4 \: – \: y_3}{x_4 \: – \: x_3} \\ = \frac{-2\: – \: (-4)}{-3\: – \: (-2)} \\ = \frac {2}{-1} \\ = \scriptsize -1\)

It is important to note that the gradient at point P is positive because the tangent \(\scriptsize \overline {PQ}\) slopes upward from left to right. Also, the gradient at point S is negative because the tangent \( \scriptsize \overline {ST}\) slopes downwards from left to right.

Zero Gradient

_{ }Given y = ax^{2} + bx + c

The graphs below show that the gradient at the turning points is zero because the tangents at points P and Q (turning points) are parallel to the x-axis.

Point P and Q are called turning points. Point P is the maximum point while point Q is the minimum point. Note that the line of symmetry in each case passes through the turning point.

### Example 1

**(a)** Copy and complete the table below

x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |

y | – | – | – | 11 | – | – | – | – |

For the value of the relation y = 11 – 2x – 2x^{2} ………for -4 â‰¤ x â‰¤ 3.

(b) Using a scale of 2cm to 1 unit on the x-axis, draw the graph of y = 11 – 2x -2x^{2}

(c) Use your graph to find:

(i) The roots of the equation 11 – 2x – 2x^{2} = 0

(ii) The value of x for which 3 – 2x – 2x^{2} = 0

(iii) The gradient of the curve at x = 1 Â Â Â Â Â Â Â **(SSCE)**

**Solution:**** **(a)

x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |

y | -13 | -1 | 7 | 11 | 11 | 7 | -1 | -13 |

(b) **Diagram of Quadratic Function Graph **

**Scale:**** **1 unit â‰¡2cm on x-axis

Â Â Â Â Â Â Â Â Â Â Â Â 5 units â‰¡ 2cm on y-axis

(c) (i) roots: x = -2.9 or x = 1.9

(ii) 3 – 2x – 2x^{2} = 0

y + (3 – 2x – 2x^{2}) = 11 – 2x – 2x^{2}

y = 11 – 2x – 2x^{2} – 3 + 2x + 2x^{2}

y = 8

values: x = 0.8 or -1.8 â‡’ y = 8

(iii) The gradient of the curve at x = 1

To find the gradient at x = 1, we will need to draw a tangent at this point as shown in the graph below.

Gradient at (x = 1) = \( \frac{y_2 \: – \: y_1}{x_2 \: – \: x_1} \\ = \frac{3.4\: – \: 7}{1.6\: – \: 1} \\ = \frac {3.6}{0.6} \\ = \scriptsize 6\)

### Example 2

(a) Copy and complete the following table for y = 2x^{2} -7x -3

x | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |

y | 19 | -3 | -9 |

(b) Using 2cm to 1unit on the x-axis and 2cm to 5units on the y-axis,

draw the graph of y = 2x^{2} – 7x – 3………..for -2 â‰¤ x â‰¤5.

From your graph find the:

(i) Minimum value of y

(ii) Gradient of the curve at x = 3 Â Â Â Â Â Â Â Â (SSCE)

**Solution:**

X | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |

Y | 19 | 6 | -3 | -8 | -9 | -6 | 1 | 12 |

**Diagram of a Quadratic Function Graph **

(i) y_{min} = -9.1

(ii) To find the gradient at x = 3, we will need to draw a tangent at this point as shown in the graph below.

Gradient = \( \frac{y_2 \: – \: y_1}{x_2 \: – \: x_1} \\ = \frac{-3 \: – \: (-6)}{3.7 \: – \: 3} \)

Gradient = \( \frac{3}{0.7} \scriptsize = 4.29 \)

â‡’ Gradient of curve (at x = 3) = 4.

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