Question

HELP ME PLZZZZZZZZZZZZZZ

Answer 1

Given:

The quadratic equation is

`y=3x^2-18x+15`

To find:

The x-coordinate and y-coordinate of the vertex.

Solution:

If a quadratic function is defined by
`f(x)=ax^2+bx+c`, then the vertex is defined as:

`Vertex=\left((-b)/(2a),f((-b)/(2a))\right)`

If
`a<0`, then the function has a maximum at the vertex and ff
`a>0`, then the function has a minimum at the vertex.

We have,

`y=3x^2-18x+15`

Here,
`a=3,b=-18,c=15`.

Since
`a=3>0`, therefore the function has a minimum at the vertex. So, fill min in first blank.

Now,

`(-b)/(2a)=(-(-18))/(2(3))`

`(-b)/(2a)=(18)/(6)`

`(-b)/(2a)=3`

Putting x=3 in the given function, we get

`y=3(3)^2-18(3)+15`

`y=27-54+15`

`y=-12`

Therefore, the vertex is at point (3,-12).