Find the x intercepts and the vertex of the parabola Objecti

Find the x intercepts and the vertex of the parabola. (Objective 2. If an answer does not exist, enter DNE.) f(x) = 4x2 48x + 128 x intercept x = c (smaller value) x intercept x = c (larger value) vertex (x, y) = , Question Part Points Submissions Used 1 2 3 4 –/1 –/1 –/1 –/1 0/10 0/10 0/10 0/10 Total –/4 . .. Find the x intercepts and the vertex of the parabola. (Objective 2. If an answer does not exist, enter DNE.) f(x) = 4x2 + 4x + 5 x intercept x = c (smaller value) x intercept x = c (larger value) vertex (x, y) = ,

Solution

The x-intercept is a point on the graph where y is zero,

Therefore, either x = (12+ 4)/2 = 16/2 = 8 or, x = (12 – 4 ) /2 = 8/2 = 4

The larger x-intercept is x = 8 and the smaller x-intercept is x = 4.

The equation of the parabola is y = 4x² -48x + 128 or, y = 4(x² – 12x + 32) = 4(x – 6)² - 4*(4) or, y = 4(x -6)² – 16. This is the vertex form of the equation of the parabola. Its vertex is (x , y) = ( 6, -16)

2. y = -4x² + 4x +5. When y =0, we have -4x² + 4x +5 = 0 or, 4x² - 4x -5 = 0. Therefore,

x = [ -(-4) ±{ ( -4)² – 4(4)(-5)} ]/ 2(4) = [4 ±( 16 + 80)] / 8 = (4±96)/ 8 =( 4 ± 46 )/ 8 = (1±6)/ 2

The larger x –intercept is x = (1+6)/2 and the smaller x –intercept is x = (1 - 6)/2.

The equation of the parabola is y = -4x² + 4x +5 = -4[ x² – 2x (1/2) + (1/4)] + 6 or, y = -4(x -1/2)² + 6.

This is the vertex form of the equation of the parabola. Its vertex is (x , y) = ( 1/2 , 6)