Graph f(x)=x^2+18x

Problem

ƒ(x)=x2+18*x

Solution

1. Identify the type of function and its basic shape. The function ƒ(x)=x2+18*x is a quadratic function in the form ƒ(x)=a*x2+b*x+c where a=1 b=18 and c=0 Since a>0 the graph is a parabola that opens upward.

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

x=−18/(2*(1))

x=−9

3. Calculate the y-coordinate of the vertex by evaluating ƒ*(−9)

ƒ*(−9)=(−9)2+18*(−9)

ƒ*(−9)=81−162

ƒ*(−9)=−81

The vertex is at (−9,−81)

4. Determine the x-intercepts by setting ƒ(x)=0 and factoring.

x2+18*x=0

x*(x+18)=0

The x-intercepts are x=0 and x=−18

5. Find the y-intercept by evaluating ƒ(0)

ƒ(0)=(0)2+18*(0)

ƒ(0)=0

The y-intercept is (0,0)

6. Sketch the graph by plotting the vertex (−9,−81) the intercepts (0,0) and (−18,0) and drawing a smooth U-shaped curve through these points.

Final Answer

ƒ(x)=x2+18*x* is a parabola with vertex *(−9,−81)* and x-intercepts *(0,0),(−18,0)

Want more problems? Check here!