Example 5: Graph y = x^2 + 2x + 3

Find the vertex and the axis of symmetry. Sketch these in.
• Find the x-intercept by plugging in 0 for y.
• Find the y-intercept by plugging in 0 for x.
• Reflect your points across the axis of symmetry and connect your dots with a smooth
U-shaped (not V-shaped) curve.

 

for now, I'm just gonna type my work and figure out what to do next

find the line of symmetry -

a = 1, b = 2, c = 3 use this to find the vertex

x = (b/2a) x = -(2) / 2(1) = -1 since we know that the along the x axis at -1 will be the vertex we replace x with 1 in the original formula

x=1  
y = x^2 -2x - 8 y = 1^2 + -2 * 1 - 8 = 1- 2 - 8 = -9 
y = -9 the vertex is (1, -9)
since the vertex is -1,-9 we know that x=1 is the axis of symmetry finding the y-intercept is the easiest to start with because we just replace x with 0 x = 0 | y = x^2 -2x - 8 y = 0 - 8 = -8 y-intercept = (0,-8) so so To find the x-intercepts, you can set y equal to zero and solve for x:

y = 0 | x = (-b ± sqrt(b^2 - 4ac)) / 2a

x = (-(-2) ± sqrt((-2)^2 - 4(1)(-8))) / 2(1)

x = (2 ± sqrt(4 + 32)) / 2

x = (2 ± sqrt(36)) / 2

x = (2 ± 6) / 2

x = 8 / 2 or x = -4/ 2

x = 4 or x = -2 

sooooo (-2,0) & (4,0) so since we know 3 y axis points on the graph and the axis of symmetry we can get another point without doing much work 

symmetry line = x = 1,  calc'd x-intercept 0,-8

the symmetry line is 1 and the known point is  0 since 1-0 = 1 we can add that to the x coordinate of y and keep the same y coordinate to get the mirrored point making another point on the graph (2,-8)  since we need one more point for the graph we can choose say x=3, | x^2 -2x - 8

y = 3^2 -3*2 - 8 = -5

soooo the new point is (3,-5) if we mirror that along 1,-9 we get  (-1, -5 ) because 3 is 2 more than 1, and 2 less than 1 is -1. we also keep the same y coordinate so all points are:

(1, -9) (0,-8) (2, -8) (3,-5) (-1,-5)