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The diagram shows part of the graph of y = x^2 - 2x + 3 (a) By drawing a suitable straight line, use your graph to find estimates for the solutions of x^2 - 3x - 1 = 0 (b) Calculate an estimate for the gradient of the graph at the point P - Edexcel - GCSE Maths - Question 20 - 2017 - Paper 2

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The diagram shows part of the graph of y = x^2 - 2x + 3 (a) By drawing a suitable straight line, use your graph to find estimates for the solutions of x^2 - 3x - ... show full transcript

Worked Solution & Example Answer:The diagram shows part of the graph of y = x^2 - 2x + 3 (a) By drawing a suitable straight line, use your graph to find estimates for the solutions of x^2 - 3x - 1 = 0 (b) Calculate an estimate for the gradient of the graph at the point P - Edexcel - GCSE Maths - Question 20 - 2017 - Paper 2

Step 1

By drawing a suitable straight line, use your graph to find estimates for the solutions of x^2 - 3x - 1 = 0

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Answer

To find the solutions for the equation x23x1=0x^2 - 3x - 1 = 0, I will first graph the function y=x23x1y = x^2 - 3x - 1 alongside the quadratic function from the graph provided. This involves plotting the points of the quadratic equation and drawing a straight horizontal line at y=0y = 0. The intersections of this line with the graph provide the estimates for the values of x. Upon careful analysis of the graph, I estimate the solutions to be approximately xextin[0.5,2.5]x ext{ in } [-0.5, 2.5].

Step 2

Calculate an estimate for the gradient of the graph at the point P.

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Answer

To estimate the gradient at the point P where x=2x=2, I first find the coordinates of point P from the graph. The corresponding y-coordinate when substituting x=2x=2 into the function y=x22x+3y = x^2 - 2x + 3 is calculated as follows:

y=222(2)+3=44+3=3y = 2^2 - 2(2) + 3 = 4 - 4 + 3 = 3

Thus, the point P is (2, 3). The next step is to determine the gradient by finding the slope of the tangent at point (2, 3). The derivative of the function

f(x)=2x2f'(x) = 2x - 2

Substituting x=2x = 2 calculates the gradient:

f(2)=2(2)2=42=2f'(2) = 2(2) - 2 = 4 - 2 = 2

Therefore, I estimate the gradient at point P to be approximately 22.

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