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

Answer:

Part A: the parabola opens upwards

Part B: $x=\frac{3}{2}$

Part C: The vertex point is $(\frac{3}{2},-\frac{1}{2})$

Part D: $y=4$

Part E: $x_{1}=1$ and $x_{2}=2$

Step-by-step 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:

$x=\frac{-b}{2a}$

$x=\frac{-(-6)}{2(2)}$

$x=\frac{6}{4}$

$x=\frac{3}{2}$

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.

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

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

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

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

$y=-\frac{1}{2}$

Therefore, the vertex point is $(\frac{3}{2},-\frac{1}{2})$

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!