Solution assignment 02 Quadratic functions and graphs

Given the function:

$y=x^2+2x-p$

Answer the following questions:

For which value of $p$ does the parabola have no intersection points with the $X$ -axis?
Calculate the intersection point with the $Y$ -axis.
Does the graph has a maximum or a minimum?

1. The parabola has no intersection points with the $X$ -axis, if $D<0$ , so:

$D=4+4p<0$
$p<-1$

2. In this case $c$ in the equation of the parabola equals $1$ and thus the intersection point with the $Y$ -axi equals $(0,1)$ .

3. Unregarded the value of $p$ we have a parabola that opens up (because $a>0$ ) and thus we have a minimum.