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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.

Screenshot 2022 06 04 at 14.27.56
y = x2 + 2x – 4

Consider the curve of y = x2 + 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. 

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 = ax2 + 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.

Screenshot 2022 06 04 at 15.14.30
Screenshot 2022 06 04 at 15.30.03

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-4-3-2-10123
y11

For the value of the relation y = 11 – 2x – 2x2 ………for  -4 ≤ x ≤ 3.

(b) Using a scale of 2cm to 1 unit on the x-axis, draw the graph of y = 11 – 2x -2x2

(c) Use your graph to find:

(i) The roots of the equation 11 – 2x – 2x2 = 0
(ii) The value of x for which 3 – 2x – 2x2 = 0
(iii) The gradient of the curve at x = 1               (SSCE)

Solution: (a) 

x-4-3-2-10123
y-13-1711117-1-13

(b) Diagram of Quadratic Function Graph 

Scale: 1 unit ≡2cm on x-axis

            5 units ≡ 2cm on y-axis

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

kofa image541

(ii) 3 – 2x – 2x2 = 0

y + (3 – 2x – 2x2) = 11 – 2x – 2x2

y = 11 – 2x – 2x2 – 3 + 2x + 2x2

y = 8

values: x = 0.8 or -1.8 ⇒ y = 8

kofa image551

(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.

kofa image56
Screenshot 2022 06 04 at 20.40.43

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 = 2x2 -7x -3

x-2-1012345
y19-3-9

(b) Using 2cm to 1unit on the x-axis and 2cm to 5units on the y-axis,
draw the graph of y = 2x2 – 7x – 3………..for -2 ≤ x ≤5.

From your graph find the:

(i) Minimum value of y
(ii) Gradient of the curve at x = 3                 (SSCE)

Solution:

X-2-1012345
Y196-3-8-9-6112

Diagram of a Quadratic Function Graph 

(i) ymin = -9.1

Screenshot 2022 06 04 at 21.03.31

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

kofa image58

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

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

⇒ Gradient of curve (at x = 3) = 4.

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