University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2023
Views:

350

Pages:

4
Author:

JackA1

IGCSE Mathematics: How to find the quadratic functions with information given IGCSE may provide you with: ● ● ● vertex and another point x-intercepts and a point vertex or intercepts with a = 1 Remember: y= a(x – h)2 + k – this is known as the vertex form of a quadratic function a is the constant (h, k) is the coordinate of the vertex Vertex forms have their uses, as you should be aware of: If a<0, we have a function with a maximum If a>0, we have a function with a minimum Important to know: When the vertex form is y = a(x + h)2 + k, The coordinate of the vertex will be (-h, k) since the value of h must be negative for the original vertex form to change: (x – (-h)) + k = (x + h) Likewise, When the vertex form is y = a(x – h)2 – k, The coordinate of the vertex will be (h, -k) since the value of k must be must be negative for the original vertex form to change: a(x – h)2 +(- k) = a(x – h)2 – k GIVEN VERTEX AND ANOTHER POINT Find the equation of the quadratic equation given the vertex (2, – 4) and the y-intercept of 8. y= a(x – h)2 + k 1st piece of information given: vertex (2, -4), therefore, h = 2 and k = -4 Substitute, y = a (x – 2)2 – 4 2nd piece of information give: y-intercept of 8, therefore, x = 0 and y = 8 Substitute, 8 = a (0 – 2)2 – 4 8 = a (-2)2 – 4 8 = 4a – 4 8 + 4 = 4a – 4 + 4 4a = 12 a=3 Therefore, the equation is y = 3(x – 2)2 – 4 or 3x2 – 12x + 8. GIVEN X-INTERCEPT AND A POINT Find the quadratic equation with x intercepts (-2, 0) and (2, 0) and passing through the point (1, 3). y= ax2 + bx + c For this question you have to find a, b and c. 1st piece of information given: x = -2, y = 0 and x = 2, y = 0 Substitute: 0 = 4a – 2b + c … [1] 0 = 4a + 2b + c … [2] [1] – [2] = 0 = -4b Therefore, b = 0 2nd piece of information given: x = 1 and y = 3 Substitute: 3=a+b+c 3 = a + c [Since b = 0] c = 3-a … [3] Substitute [3] into [1] 0 = 4a + 3 – a 3a + 3 = 0 Therefore, a = -1 Since a = -1 and b = 0, We can substitute this information to [1] to find c, 0 = -4 + c Hence, c = 4 Consider y= ax2 + bx + c, y = -x2 + 4 VERTEX OR INTERCEPTS WITH A = 1 Find the quadratic equation in the form y = a(x – b)2 + c When the vertex is at (3, -1) and the value of a = 1. Information given: a = 1, h = 3 and k = -1 Substitute into y= a(x – h)2 + k: y = (x – 3)2 – 1 y = x2 – 6x + 8