What are the x-intercepts of y = 2x² - 3x - 20?

Solution:

The given function y = 2x² - 3x - 20 is a quadratic function and represents a parabola facing upwards because coefficient of x² is positive.

Its vertex is given by the x-coordinate of :

x = -b/2a

Here b = -3 and a = 2

Therefore

x = -(-3)/(2)(2) = 3/4

Therefore y coordinate of the vertex is

y = 2(3/4)² - 3(3/4) - 20

= 2(9/16) - 9/4 - 20

= 9/8 - 9/4 - 20

= -9/8 - 20

= -169/20

The vertex is (3/4, -169/20) or (0.75, -8.45)

The x intercepts can be determined by either factorizing the given equation or determining its roots:

y = 2x² - 3x - 20

y = 2x² - 8x + 5x - 20

y = 2x(x - 4) + 5(x - 4)

= (x - 4)(2x + 5)

To determine the zeros of the function we put y = 0 and therefore

(x - 4)(2x + 5) = 0

Hence (x- 4) = 0 and (2x + 5) = 0

x - 4 = 0 ⇒ x = 4

2x + 5 = 0 ⇒ x = -5/2

Thus the x-intercepts are (4,0) and  (-5/2, 0)

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What are the x-intercepts of y = 2x² - 3x - 20?

Summary:

The x-intercepts of y = 2x² - 3x - 20 are (4,0) and  (-5/2, 0)