Question

I need help with finding the vertex,range and domain interval

Answer 1

THis is the graph of a positive parabola, that opens upward.

The equation is:

$y=x^2+2x-3$

The general form of a quadratic is >>>

$ax^2+bx+c$

So,

a = 1

b = 2

c = -3

(a)

The vertex is the lowest point of the parabola. The x-coordinate is:

$\begin{gathered} x=-\frac{b}{2a} \\ x=-\frac{2}{2(1)} \\ x=-1 \end{gathered}$

And y is:

$\begin{gathered} y=x^2+2x-3 \\ y=(-1)^2+2(-1)-3 \\ y=1-2-3 \\ y=-4 \end{gathered}$

The vertex is at (-1, -4).

(b)

The range is the set of y-values for which a function is defined. From the graph, we can see that the parabola is defined from y = -4 to y = 0. In interval notation >>>

$\text{Range}=\lbrack-4,0\rbrack$(c)

The domain is the set of x-values for which a function is defined.

The two x-axis points are the x-intercepts. They are found below:

$\begin{gathered} y=x^2+2x-3 \\ y=(x+3)(x-1) \\ 0=(x+3)(x-1) \\ x=-3,1 \end{gathered}$

From the graph, we can see that the parabola is defined from x = -3 (not included point) until x = 1. So, the domain is >>>

$\text{Domain}=(-3,1\rbrack$