Graph y=x^2+2x+5

Problem

y=x2+2*x+5

Solution

1. Identify the type of function. This is a quadratic function in the form y=a*x2+b*x+c where a=1 b=2 and c=5 Since a>0 the parabola opens upward.

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

x=−2/(2*(1))=−1

3. Calculate the y-coordinate of the vertex by substituting x=−1 into the original equation.

y=(−1)2+2*(−1)+5=1−2+5=4

The vertex is at (−1,4)

4. Find the y-intercept by setting x=0

y=0+2*(0)+5=5

The y-intercept is at (0,5)

5. Determine the x-intercepts by setting y=0 and checking the discriminant D=b2−4*a*c

D=2−4*(1)*(5)=4−20=−16

Since the discriminant is negative, there are no real x-intercepts.

6. Find additional points to improve the graph's accuracy. For x=−2

y=(−2)2+2*(−2)+5=4−4+5=5

The point is (−2,5)

7. Sketch the graph by plotting the vertex (−1,4) the y-intercept (0,5) and the symmetric point (−2,5) then drawing a smooth upward-opening curve through them.

Final Answer

y=x2+2*x+5

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