Graph x^2-2x

Problem

y=x2−2*x

Solution

1. Identify the function type and its coefficients. This is a quadratic function in the form y=a*x2+b*x+c where a=1 b=−2 and c=0 Since a>0 the parabola opens upward.

2. Find the x-intercepts by setting y=0 and factoring the expression.

x2−2*x=0

x*(x−2)=0

The intercepts are at x=0 and x=2

3. Determine the vertex using the formula x=(−b)/(2*a)

x=(−(−2))/(2*(1))=1

Substitute x=1 into the original equation to find the y-coordinate.

y=(1)2−2*(1)=−1

The vertex is at (1,−1)

4. Identify the axis of symmetry, which is the vertical line passing through the vertex.

x=1

5. Find the y-intercept by evaluating the function at x=0

y=(0)2−2*(0)=0

The y-intercept is (0,0)

6. Plot the points and draw a smooth curve. Use the vertex (1,−1) the x-intercepts (0,0) and (2,0) and additional points like (3,3) and (−1,3) to define the shape.

Final Answer

y=x2−2*x

Want more problems? Check here!