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:

$Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)$

If $a$, 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,

$\dfrac{-b}{2a}=\dfrac{-(-18)}{2(3)}$

$\dfrac{-b}{2a}=\dfrac{18}{6}$

$\dfrac{-b}{2a}=3$

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).