Question

What are the x and y intercepts for `y = 1/2 (x-4)^2 +18`?

Answer 1

There is no `x`-intercept. `y`-intercept is `26`.

Explanation:

To find `x`-intercept of any curve, just put `y=0`

and to `x`-intercept of any curve, just put `x=0`.

Hence `x`-intercept of `y=1/2(x-4)^2+18` is given by `1/2(x-4)^2+18=0` or `1/2(x-4)^2=-18`. But this is not possible asLHS cannot be negative. Hence, we do not have `x`-intercept.

For `y`-intercept of `y=1/2(x-4)^2+18`, put `x=0` and then `y=1/2*(-4)^2+18=26`. Hence `y`-intercept is `26`.

graph{y=1/2(x-4)^2+18 [-77, 83, -18.56, 61.44]}