Graph f(x)=3(x+2)^2-7

Problem

ƒ(x)=3*(x+2)^2−7

Solution

1. Identify the form of the function. The equation is in vertex form, ƒ(x)=a*(x−h)^2+k where (h,k) is the vertex of the parabola.

2. Determine the vertex by identifying h and k Since the expression is 3*(x−(−2))^2−7 the vertex is (−2,−7)

3. Determine the direction of the opening. Since the lead coefficient a=3 is positive, the parabola opens upward.

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

ƒ(0)=3*(0+2)^2−7

ƒ(0)=3*(4)−7

ƒ(0)=5

The y-intercept is (0,5)

5. Find additional points using symmetry. Since the axis of symmetry is x=−2 the point symmetric to the y-intercept (0,5) is (−4,5)

6. Plot the points and draw a smooth U-shaped curve through the vertex (−2,−7) the y-intercept (0,5) and the symmetric point (−4,5)

Final Answer

To graph ƒ(x)=3*(x+2)^2−7 plot the vertex at (−2,−7) the y-intercept at (0,5) and use the axis of symmetry x=−2 to draw an upward-opening parabola.

Want more problems? Check here!