\(\begin{array}{c|c} x & 0 & 1\frac{1}{4} & 2 & 4\\ \hline y & 3 & 5\frac{1}{2} & & \end{array}\) The table given shows some values for a linear graph. Find...

Assessment: WAEC SSCE - General Mathematics - 2015 (Objective) Subject: General Mathematics

Question 1 Report

\(\begin{array}{c|c}
x & 0 & 1\frac{1}{4} & 2 & 4\\
\hline
y & 3 & 5\frac{1}{2} & &
\end{array}\)
The table given shows some values for a linear graph. Find the gradient of the line

Answer Details
To find the gradient of a line, we need to calculate the change in y divided by the change in x between two points on the line. Let's choose two points on the line from the given table. We can select the points (0,3) and (2,y), where y is the value of y for x = 2. The change in y is y - 3, and the change in x is 2 - 0 = 2. Therefore, the gradient of the line is: gradient = change in y / change in x = (y - 3) / 2 We don't know the value of y yet, but we can use the other point on the line to find it. Using the points (1.25, 5.5) and (2, y), we get: gradient = (y - 5.5) / (2 - 1.25) = (y - 5.5) / 0.75 Setting the two expressions for the gradient equal to each other, we get: (y - 3) / 2 = (y - 5.5) / 0.75 Solving for y gives: y = 7 Therefore, the value of y for x = 2 is 7, and the gradient of the line is: gradient = (y - 3) / 2 = (7 - 3) / 2 = 2 So the answer is 2, and the gradient of the line is 2. In simple terms, the gradient of a line tells us how steep the line is. If the gradient is positive, the line is sloping upwards from left to right. If the gradient is negative, the line is sloping downwards from left to right. The larger the absolute value of the gradient, the steeper the line.

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