Parabolas

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Math › Parabolas

Questions 1 - 4

1

Find the vertex $\left ( x,y \right )$ for a parabola with equation

$\left ( 1,-2 \right )$

$\left ( 1,1 \right )$

$\left ( 1,2 \right )$

$\left ( 2,-1 \right )$

$\left ( 1,3 \right )$

Explanation

For any parabola of the form , the -coordinate of its vertex is

$x = \frac{-b}{2a}$

So here, we have

$x=\frac{-(-6))}{2(3)}$

=

We plug this back into the original equation to find :

=

2

Find the vertex $\left ( x,y \right )$ for a parabola with equation

$\left ( 1,-2 \right )$

$\left ( 1,1 \right )$

$\left ( 1,2 \right )$

$\left ( 2,-1 \right )$

$\left ( 1,3 \right )$

Explanation

For any parabola of the form , the -coordinate of its vertex is

$x = \frac{-b}{2a}$

So here, we have

$x=\frac{-(-6))}{2(3)}$

=

We plug this back into the original equation to find :

=

3

What is the minimal value of

over all real numbers?

No minimum value.

Explanation

Since this is an upwards-opening parabola, its minimum value will occur at the vertex. The -coordinate for the vertex of any parabola of the form

is at

$x = \frac{-b}{2a}$

So here,

$x = \frac{-b}{2a}$

$= \frac{-(16)}{2(2)}$

We plug this value back into the equation of the parabola, to find the value of the function at this .

Thus the minimal value of the expression is .

4

What is the minimal value of

over all real numbers?

No minimum value.

Explanation

Since this is an upwards-opening parabola, its minimum value will occur at the vertex. The -coordinate for the vertex of any parabola of the form

is at

$x = \frac{-b}{2a}$

So here,

$x = \frac{-b}{2a}$

$= \frac{-(16)}{2(2)}$

We plug this value back into the equation of the parabola, to find the value of the function at this .

Thus the minimal value of the expression is .

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