Answer
See graph
Work Step by Step
We are given the function:
$f(x)=x^2+4x-5$
Determine its coefficients from the standard form:
$f(x)=ax^2+bx+c$
$a=1$
$b=4$
$c=-5$
Because $a>0$ the graph opens up.
Determine the vertex $V(x_V,y_V)$:
$x_V=-\dfrac{b}{2a}=-\dfrac{4}{2(1)}=-2$
$y_V=-\dfrac{b^2-4ac}{4a}=-\dfrac{4^2-4(1)(-5)}{4(1)}=-9$
$V(-2,-9)$
The axis of symmetry is:
$x=x_V$
$x=-2$
Determine the intercepts:
$x^2+4x-5=0$
$x=\dfrac{-4\pm\sqrt{4^2-4(1)(-5)}}{2(1)}=\dfrac{-4\pm 6}{2}=-2\pm 3$
$x_1=-2-3=-5$
$x_2=-2+3=1$
$(-5,0),(1,0)$
Plot the vertex, the intercepts, and the axis of symmetry. Then graph the function:
