Question

Find the maximum or minimum of the following quadratic function: y=x^2 -16

Answer 1

so.... any idesa where the vertex of that one is at?

Answer 2

\[x^{2}\] It's supposed to look like that.

Answer 3

I think the vertex is X.

Answer 4

well... have you covered quadratics yet?

Answer 5

Yes, but it's not my strong point.

Answer 6

well let's take a peek ....
\(\large {
y=(x-{\color{brown}{ h}})^2+{\color{blue}{ k}}\\
x=(y-{\color{blue}{ k}})^2+{\color{brown}{ h}}\qquad\qquad vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})
}\)

notice that, notice where the vertex is at

Answer 7

\(\bf y=x^2-16\implies y-{\color{blue}{ 0}}=(x-{\color{brown}{ 0}})^2-16\)

notice the vertext on that one

and the -16 is just a transformation of \(x^2\)

-16 means, from that vertex, you move down 16 units

Answer 8

It's at K?

Answer 9

well... actually. hmmm hold the mayo... I'm mixing up two forms there...

Answer 10

yes, just use the "k" form the vertex form

Answer 11

lemme rewrite that quick

Answer 12

\(\bf y=x^2-16\implies y=(x-{\color{brown}{ 0}})^2{\color{blue}{ -16}}\)

there, now you can see where the vertex is at :)

Answer 13

So the vertex is sixteen?

Answer 14

the \(x^2\ has a positive coefficient
meaning the parabola goes upwards

a maximum point means a "hump"
a minimum point, means a "burrow"

in this case the parabola has a "burrow" not a "hump"
thus it has a minumum then