For each of the following quadratic functions find its opening, vertex, axis of symmetry, ymin or y max, y-intercept, x-intercept

Question

(c) y=3x²+6x+2

(d) y=8x-5-x²

Answer 1

y=3x^2+6x+2

opens up

vertex: -b/2a=-6/6=-1

x=-1, y=3(1)+6(-1)+2=3-6+2=-1

y=-1

vertex=(-1,-1)

axis of symmetry: x=-b/2a=-6/6=-1

axis of symmetry: x=-1

the parabola opens up so there is a min, the y-value of the vertex

min=-1

to find the y-intercept set x=0 and find y

y=2 is the y-intercept

to find the x-intercept set y=0 and find x (x's)

0=3x^2+6x+2

use the quadratic equation

(-6±√[36-24])/6

x=(-6±√[12])/6

x=-1±√3/3 because √12=2√3

y=8x-5-x^2

y=-x^2+8x-5

opens down

vertex:-8/-2=4

x=4

y=-16+32-5=11

vertex:(4,11)

axis of symmetry: x=4

max=11, the y-value of the vertex

y-intercept:set x=0 and y=-5

x-intercept: set y=0

0=-x^2+8x-5

x^2-8x+5=0

use the quadratic formula

(8±√[64-20])/2=

(8±√44)/2=

(8±2√11)/2=

4±√11