Question

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

Answer 1

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!