Question

Graph the parabola.Y = 2x^2 - 4x + 5Plat five points on a parable the vertex ,two points to the left of the vertex ,and two points to the right of the vertex . then click on the graph a function button.

Answer 1

Given a function to graph:

`y=2x^2-4x+5`

First, we have to the vertex of the parabola.

We know that the vertex of the parabola is at x = -b/2a. For the given equation, a = 2, b = -4, c = 5. So, the vertex is:

`\begin{gathered} x=(-b)/(2a) \\ x=(-(-4))/(2(2)) \\ x=(4)/(4) \\ x=1 \end{gathered}`

Now, at x = 1, the value of y is:

`\begin{gathered} y=2(1)^2-4(1)+5 \\ y=2-4+5 \\ y=7-4 \\ y=3 \end{gathered}`

So, the vertex of the given parabola is (1, 3).

The two points to the left of the vertex can be obtained at x = 0, -1.

At x = 0.

`\begin{gathered} y=2(0)^2-4(0)+5 \\ y=5 \end{gathered}`

At x = -1.

`\begin{gathered} y=2(-1)^2-4(-1)+5 \\ =2+4+5 \\ =11 \end{gathered}`

So, the two points to the left of the vertex are (0, 5) and (-1, 11).

The two points to the right of the vertex can be obtained at x = 2 and x = 3.

At x = 2.

`\begin{gathered} y=2(2)^2-4(2)+5 \\ =8-8+5 \\ =5 \end{gathered}`

At x = 3.

`\begin{gathered} y=2(3)^2-4(3)+5 \\ y=18-12+5 \\ y=11 \end{gathered}`

So, the two points to the right of the vertex are (2, 5) and (3, 11).

Plot these points on the graph to get the graph as follows: