117k views
24 votes
Joel was studying the function f(x) = 2x2– 6x + 4. Please identify the key points. Part A: Direction of opening Part B: Axis of Symmetry Part C: Vertex Part D: Y-intercept Part E: X-intercept

1 Answer

8 votes

Answer:

Part A: the parabola opens upwards

Part B:
x=(3)/(2)

Part C: The vertex point is
((3)/(2),-(1)/(2))

Part D:
y=4

Part E:
x_(1)=1 and
x_(2)=2

Explanation:

We have the function:


f(x)=2x^(2)-6x+4

Where the coefficients are:

a = 2

b = -6

c = 4

Part A:

Here we can see that the leading coefficient of our parabola (a = 2) is positive, so by the definition, the parabola opens upwards.

Part B:

The axis of the symmetry equation is given by:


x=(-b)/(2a)


x=(-(-6))/(2(2))


x=(6)/(4)


x=(3)/(2)

Part C:

The vertex is the minimum point of our parabola.

Using the x value founded in Part B we can find f(x)=y.


y=2(3/2)^(2)-6(3/2)+4


y=2(9/4)-6(3/2)+4


y=(9/2)-3(3)+4


y=(9/2)-9+4


y=-(1)/(2)

Therefore, the vertex point is
((3)/(2),-(1)/(2))

Part D:

To get the y-intercept we just need to do x = 0.


y=2(0)^(2)-6(0)+4


y=4

The y-intercept is (0,4)

Part E:

To get the x-intercept we just need to do y = 0.


0=2x^(2)-6x+4


2(x-2)(x-1)=0


x_(1)=1


x_(2)=2

The x-intercept is (1,0) and (2,0)

I hope it helps you!