WRITING QUADRATIC EQUATIONS IN INTERCEPT FORM

Find the zeros of the function by rewriting the function in intercept form.

Problem 1 :

y = 3x² + 2x

Solution :

Finding x-intercepts :

Put y = 0

 3x² + 2x = 0

x(3x + 2) = 0

3x + 2 = 0

3x = -2

x = -2/3 or x = 0

Intercept form f(x) = x (3x + 2)

Problem 2 :

y = 12x² + 8x - 15

Solution :

Finding x-intercepts :

Put y = 0

 12x² + 8x - 15 = 0

 12x² + 18x - 10x - 15 = 0

 6x (2x + 3) - 5(2x + 3) = 0

 (6x - 5) (2x + 3) = 0

6x - 5 = 0     2x + 3 = 0

6x = 5    2x = -3

x = 5/6    x = -3/2

Intercept form f(x) = (6x - 5) (2x + 3)

Problem 3 :

f(x) = 5x² - 25x + 30

Solution :

Finding x-intercepts :

Put y = 0

5x² - 25x + 30 = 0

5(x² - 5x + 6) = 0

5(x² - 2x - 3x + 6) = 0

5[x(x - 2) - 3(x - 2)] = 0

5[(x - 2) (x - 3)] = 0

x - 2 = 0 or x - 3 = 0

x = 2 or x = 3

Intercept form f(x) = 5 (x - 2) (x - 3)

Problem 4 :

y = 25x² + 10x - 24

Solution :

Finding x-intercepts :

Put y = 0

25x² + 10x - 24 = 0

25x² + 30x - 20x - 24 = 0

5x(5x + 6) - 4(5x + 6) = 0

(5x - 4) (5x + 6) = 0

5x - 4 = 0 or 5x + 6 = 0

5x = 4 or 5x = -6

x = 4/5 or x = -6/5

Intercept form f(x) = (5x - 4) (5x + 6)

Problem 5 :

g(x) = 33x² - 9x - 24

Solution :

Finding x-intercepts :

Put y = 0

33x² - 9x - 24 = 0

3(11x² - 3x - 8) = 0

3(11x² + 8x - 11x - 8) = 0

3[x (11x + 8) - 1(11x + 8)] = 0

3(x - 1) (11x + 8) = 0

x - 1 = 0 or 11x + 8 = 0

x = 1 or 11x = -8

      x = -8/11

Intercept form f(x) = 3(x - 1) (11x + 8)

Problem 6 :

y = 4x² + 1

Solution :

4x² + 1= 0

The given quadratic polynomial is not factorable. While solving it,

4x² = - 1

x² = -1/4

x = √(-1/4)