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|x^2+x-2|
Plotting a function graph/ |x^2+x-2|

Graphing y = |x^2+x-2|

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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       | 2        |
f(x) = |x  + x - 2|
f(x)=∣x2+x−2∣f{\left(x \right)} = \left|{x^{2} + x - 2}\right| 
f = |x^2 + x - 1*2|
The graph of the function
02468-8-6-4-2-10100200
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis X at f = 0
so we need to solve the equation:
∣x2+x−2∣=0\left|{x^{2} + x - 2}\right| = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−2x_{1} = -2
x2=1x_{2} = 1
Numerical solution
x1=−2x_{1} = -2
x2=1x_{2} = 1
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to |x^2 + x - 1*2|.
∣(−1)2+02+0∣\left|{\left(-1\right) 2 + 0^{2} + 0}\right|
The result:
f(0)=2f{\left(0 \right)} = 2
The point:
(0, 2)
Inflection points
Let's find the inflection points, we'll need to solve the equation for this
d2dx2f(x)=0\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
d2dx2f(x)=\frac{d^{2}}{d x^{2}} f{\left(x \right)} =
the second derivative
2((2x+1)2δ(x2+x−2)+sign⁡(x2+x−2))=02 \left(\left(2 x + 1\right)^{2} \delta\left(x^{2} + x - 2\right) + \operatorname{sign}{\left(x^{2} + x - 2 \right)}\right) = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞∣x2+x−2∣=∞\lim_{x \to -\infty} \left|{x^{2} + x - 2}\right| = \infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞∣x2+x−2∣=∞\lim_{x \to \infty} \left|{x^{2} + x - 2}\right| = \infty
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of |x^2 + x - 1*2|, divided by x at x->+oo and x ->-oo
lim⁡x→−∞(∣x2+x−2∣x)=−∞\lim_{x \to -\infty}\left(\frac{\left|{x^{2} + x - 2}\right|}{x}\right) = -\infty
Let's take the limit
so,
inclined asymptote on the left doesn’t exist
lim⁡x→∞(∣x2+x−2∣x)=∞\lim_{x \to \infty}\left(\frac{\left|{x^{2} + x - 2}\right|}{x}\right) = \infty
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
Even and odd functions
Let's check, whether the function even or odd by using relations f = f(-x) и f = -f(-x).
So, check:
∣x2+x−2∣=∣−x2+x+2∣\left|{x^{2} + x - 2}\right| = \left|{- x^{2} + x + 2}\right|
- No
∣x2+x−2∣=−∣−x2+x+2∣\left|{x^{2} + x - 2}\right| = - \left|{- x^{2} + x + 2}\right|
- No
so, the function
not is
neither even, nor odd
The graph
Graphing y = |x^2+x-2|
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