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Question

Consider f(x) = 3x^2 - 12x + 4.
find the vertex, intercepts, max/min value of the function, and graph it.

Answer 1

The given fuction f(x) = 3x²-12x+4

The above equation compare to parabola equation  y = ax²+bx+c

Hence a = 3, b = -12 and c = 4

The maximum or minimum value of the quadratic function at x = -(b/2a)

x = -(-12)/2.3 = 2

To find the minimum value of the function, which is also the y-value f(2)

f(2) = 3(2)²-12(2)+4

y = 16 - 24 = -8

The minimum value of the fuction is x = 2  and y = -8 and the vertex = ( 2,-8)

we solve f(x) = 0 for know the x-intercepts

f(x) = 3x²-12x+4

Add  -4 to each side

3x²-12x+4 -4 = -4

3(x²-12x+4-4) = -4

3(x -2)²- 12 = -4

3(x -2)²= -4+12

(x -2)²= 8/3

Take square each side

(x -2)  = √8/3

Therefore x = (± 1.63)+2

Therefore the intercept  values is (3.66, 0.37)

 Graph of the fuction is 

 

Comments

The function f(x) = 3x²-12x + 4.

To find the y - intercept, substitute the value of x = 0 in the original equation.

f(0) = 3(0)²-12(0) + 4 = 4.

Therefore, y - intercept is 4.