I'm doing my math work and I have to sketch the description of a polynomial function.
The question I'm doing states that this graph has
two non-real zeros
I was wondering what this looks like, and if non-real numbers refer to any complex number with i?
Thank you :)
Step back.
If a graph has no real zeros[*] then the graph never has values where $y=0$ and therefore the graph never crosses the $x$-axis.
And that's your answer in a single sentence. This looks like a graph that never crosses the $x$ axis and is thus either entirely above or below the $x$ axis.
More thoughtful:
A quadratic polynomial $f(x) = ax^2 + bx + c$ will have two solutions $x = \frac {-b +\sqrt{b^2 -4ac}}2$ and $x = \frac {-b-\sqrt{b^2 -4ac}}2$. So long as $b^2 -4ac > 0$ then these solutions will have real roots. (If $b^2 -4ac = 0$ these roots will both equal each other).
Remember that the graph of a parabola: has a cusp at $x =\frac {-b}{2a}, y = c-\frac {b^2}{4a}$. If $a > 0$ and $b^2 > 4ac$ then the parabola points up and $y = c-\frac {b^2}{4a} < 0$ and the cusp is below the $x$ axis and there are two zeros.
But if $a > 0$ and $b^2 < 4ac$ then $y = c-\frac {b^2}{4a} > 0$ and the parabola points up and the cusp is above the $x$ axis and there are no zeros.
And if $a > 0$ and $b^2 =4ac$ then $y = c-\frac {b^2}{4a}=0$ and the cusp is on zero making a single root.
(If $a < 0$ then similar result but the parabola is pointing down)
.....
A simple example would by that $y = x^2 + 1> 1 > 0$ so it never has any roots. It is strictly above the $x$ axis. If we tried to solve $x^2 + 1=0$ so $x^2 = -1$ there.... are no real roots! C'est tout!
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[*] I assuming when the say two non-real roots they mean zero real roots. It's possible for there two be a function with two non-real roots and some real roots.
Ex: $x^2 + 1$ has no real roots and $x+2$ has one real root (at $x = -2$) so $f(x) = (x^2 + 1)(x+2) = x^3+2x^2 + x + 2$ will have one real root at $-2$ and two non-real roots (at $i$ and $-i$ if you are curious). The graph will look like a usual 3rd degree "snake bend" but instead of crossing the $x$ axis and bending down crossing it again and bending up and crossing it a third time, it will cross the $x$ axis once, then bend down but not go down enough to cross and then will bend up before it ever crosses and as a result will only cross the $x$ axis once.