The diagram shows part of the graph of a quadratic function, with equation of the form y=k(x-a)(x-b). The graph cuts the v-axis at (0,-6) and the x-axis at (-1, 0) and (3, 0)
a. Write down the values of a and b.
b. Calculate the value of k.
c. Identify the coordinates of the minimum turning point of the function.
Answers
Given : part of the graph of a quadratic function, with equation of the form y=k(x-a)(x-b).
The graph cuts the y-axis at (0,-6) and the x-axis at (-1, 0) and (3, 0)
To Find : a. Write down the values of a and b.
b. Calculate the value of k.
c. Identify the coordinates of the minimum turning point of the function.
Solution:
y=k(x-a)(x-b)
the x-axis at (-1, 0) and (3, 0)
a = - 1
b = 0
values of a and b -1 and 0
y = k (x-(-1))(x - 3)
=> y = k(x + 1)(x - 3)
The graph cuts the y-axis at (0,-6)
=> -6 = k(0 + 1)(0 - 3)
=> -6 = k (1)(-3)
=> k = 2
value of k. = 2
y = 2(x + 1)(x - 3)
dy/dx = 2(x +1) + 2(x - 3) = 4x - 4
4x - 4 = 0
=> x = 1
d²y/dx² = 4 > 0 hence minimum
y = 2(x + 1)(x - 3)
x = 1
=> y = 2(1 + 1)(1 - 3) = - 8
( 1 , - 8) the coordinates of the minimum turning point of the function
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