FIND QUADRATIC EQUATION IN STANDARD FORM FROM THE GIVEN INFORMATION

Find, in the form y = ax2 + bx + c, the equation of the quadratic whose graph:

Problem 1 :

cuts the x-axis at 5 and 1, and passes through (2, -9).

Solution:

y = k(x - 5)(x - 1) where k ≠ 0

It passes through (2, -9). Substitute (x, y) = (2, -9)

-9 = k(2 - 5)(2 - 1)

-9 = k(-3)(1)

-9 = -3k

k = 3

By applying k = 3 in above equation

y = 3(x - 5)(x - 1)

y = 3(x2 - x - 5x + 5)

y = 3(x2 - 6x + 5)

y = 3x2 - 18x + 15

Problem 2 :

cuts the x-axis at 2 and -1/2, and passes through (3, -14)

Solution:

y=k(x-2)x+12

It passes through (3, -14). Substitute (x, y) = (3, -14)

-14=k(3-2)3+12-14=k(1)72-14=72kk=-14×27k=-4

By applying k = -4 in above equation

y=-4(x-2)x+12y=-4x2+12x-2x-1y=-4x2-32x-1y=-4x2+6x+4

Problem 3 :

touches the x-axis at 3 and passes through (-2, -25)

Solution:

Since the x- intercept is 3(repeated), the equation is

y = k(x - 3)2 where k ≠ 0

It passes through (-2, -25). Substitute (x, y) = (-2, -25)

-25 = k(-2 - 3)2 

-25 = k(-5)2

25k = -25

k = -1

By applying k = -1 in above equation

y = -1(x - 3)²

y = -(x² - 6x + 9)

y = -x² + 6x - 9

Problem 4 :

touches the x-axis at -2 and passes through (-1, 4)

Solution:

Since the x- intercept is -2(repeated), the equation is

y = k(x + 2)2  

It passes through (-1, 4). Substitute (x, y) = (-1, 4)

4 = k(-1 + 2)2 

4 = k(1)2

k = 4

By applying k = 4 in above equation

y = 4(x + 2)²

y = 4(x2 + 4x + 4)

y = 4x2 + 16x + 16

Problem 5 :

cuts the x-axis at 3, passes through (5, 12) and has axis of symmetry x = 2

Solution:

Since, the x- intercept is 3 and the axis of symmetry is x = 2, the other x- intercept is 1, the equation is

y = k(x - 1)(x - 3) where k ≠ 0

It passes through (5, 12). Substitute (x, y) = (5, 12)

12 = k(5 - 1)(5 - 3)

12 = k(4)(2)

12 = 8k

k = 3/2

By applying k = 3/2 in above equation

y=32(x-1)(x-3)y=32x2-3x-x+3y=32x2-4x+3y=32x2-6x+92

Problem 6 :

cuts the x-axis at 5, passes through (2, 5) and has axis of symmetry x = 1

Solution:

Since, the x- intercept is 5 and the axis of symmetry is x = 1, the other x- intercept is -3, the equation is

y = k(x - 5)(x + 3) where k ≠ 0

It passes through (2, 5). Substitute (x, y) = (2, 5)

5 = k(2 - 5)(2 + 3)

5 = k(-3)(5)

5 = -15k

k = -1/3

By applying k = -1/3 in above equation

y=-13(x-5)(x+3)y=-13x2+3x-5x-15y=-13x2-2x-15y=-13x2+23x+5

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