Question

y=(x−5)^2−4
Identify the vertex, equation for the axis of symmetry, domain, range and x- and y-intercepts of the parabola.


y=−2(x+8)^2+2
Identify the vertex, equation for the axis of symmetry, domain, range and x- and y-intercepts of the parabola.

Answer 1

Compare it to the vertex form of the parabola y = a(x-h)^2 + k
where (h,k) is the vertex and you can quickly identify the vertex.

Answer 2

how about we get married? :DDD

Answer 3

y=(x−5)^2−4 is an vertical parabola. Therefore the axis of symmetry will be a vertical line with the equation x = h where h ix the x-coordinate of the vertex found earlier.

Answer 4

vertex = -b/2a
domain is = x
range = y
x intercept is when y is = 0
y intercept is when x is = 0

Answer 5

i got the vertex, axis of symmetry, domain, and range but i cant figure out the intercepts

Answer 6

or the vertex can be (h,k) y = x - h)^2 + k

Answer 7

To find x-intercept, set y = 0 and solve for x.
To find y-intercept, set x = 0 and solve for y.

Answer 8

you have this equation y=(x−5)^2−4
if you want to find the y intercept you just have to plug in the zero for the x
y=((0)-5)^2 - 4
if you want to find the x intercept you just have to plug in thr zero for the y
(0)=(x-5)^2-4

Answer 9

so y=21 but for im not exactly getting a real answer for x

Answer 10

i got x^2-10x-21=0

Answer 11

(0)=(x-5)^2-4
4 = (x-5)^2
take square root
+/- 2 = x-5
+2 = x - 5 gives x = 7
-2 = x - 5 gives x = 3

Answer 12

you can just do this 0=(x-5)^2 - 4
4 = (x-5)^2
sqrt(4) = x - 5
2 = x - 5
x =7

Answer 13

y-intercept: 21
x-intercepts: 3, 7

Answer 14

or what aum wrote

Answer 15

thank youu