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For the graph of f(x)=2x28x+1

Find:

鈸 the axis of symmetry

鈸 the vertex.

Short Answer

Expert verified

鈸 The axis of symmetry is the line x = 2.

鈸 The vertex is (2, 鈭7) .

Step by step solution

01

Step 1. Given Information

For the graph of f(x)=2x28x+1

We want to find:

鈸 the axis of symmetry

鈸 the vertex.

We know that Axis of Symmetry and Vertex of a Parabola:

The graph of the function f(x)=ax2+bx+cis a parabola where:

  • the axis of symmetry is the vertical linex=b2a
  • the vertex is a point on the axis of symmetry, so its x-coordinate isb2a
  • the y-coordinate of the vertex is found by substituting x=b2ainto the quadratic equation.
02

Part (a) Step 1. Solve for the axis of symmetry

Solution of the axis of symmetry

f(x)=ax2+bx+cf(x)=2x28x+1

The axis of symmetry is the vertical line

x=b2a

Substitute the values of a, b into the

equation.

localid="1646149690675" x=--82(2)

Simplify

x=2

The axis of symmetry is the line x = 2 .

03

Part (b) Step 1. Solve for the vertex

f(x)=ax2+bx+cf(x)=2x28x+1

The vertex is a point on the line of

symmetry, so its x-coordinate will be

x = 2 .

Find f (2) .

f(x)=2x28x+1f(2)=2(2)28(2)+1f(2)=816+1f(2)=-7

The vertex is (2, 鈭7) .

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