Question

1. Graph each of the following equations below using a tale of values or by another method. Fill in theInformation for each graph.X-intercept:a) y = x2 + 4x - 5y-intercept:хуVertex:Max/MinAxis of SymmetryDomain:Range:

Answer 1

The equation is

`y=x^2+4x-5`

We can already find the vertex using the vertex formulas

`\begin{gathered} x_V=-(b)/(2a)=(-4)/(2)=-2 \\ \\ \\ y_V=-(\Delta)/(4a)=-(b^2-4ac)/(4a)=-(16+20)/(4)=-(36)/(4)=-9 \end{gathered}`

Therefore the vertex is

`(x_V,y_V)=(-2,-9)`

Now we have the vertex we also have the axis of symmetry and the max/min of the function, in that case, it's a minimum because a > 0. Therefore

`\begin{gathered} \text{ axis of symmetry = }x_V=-2 \\ \\ \min\lbrace y\rbrace=y_V=-9 \end{gathered}`

We can find the x-intercept easily

`\begin{gathered} x=(-b\pm√(b^2-4ac))/(2a) \\ \\ x=(-4\pm√(4^2+4\cdot5))/(2) \\ \\ x=(-4\pm√(16+20))/(2) \\ \\ x=(-4\pm√(36))/(2) \\ \\ x=(-4\pm6)/(2) \\ \\ \end{gathered}`

Hence

`\begin{gathered} x=(-4\pm6)/(2)=-2\pm3 \\ \\ x_1=-1 \\ x_2=-5 \end{gathered}`

The y-intercept is just the c value, then it's -5.

Now we can do the domain, there's no restriction for parabolas in the domain, then

`\text{ domain = }\mathbb{R}`

And the range is

`\text{ range = \lbrack}y_V,+\infty)=[-9,+\infty)`