How do you use the important points to sketch the graph of $f (x) = 3 - x^{2} - 2 x$ $f(x)=3-x^2-2x$ ?

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Answer 1 · 2016-09-17T11:12:32

$f (x)$ $f(x)$ has zeros at $x = - 3 and 1$ $x=-3 and 1$
$f (x)$ $f(x)$ has a maximum at $(- 1, 4)$ $(-1,4)$

$f (x) = 3 - x^{2} - 2 x$ $f(x) = 3-x^2-2x$
$f (x) = 0 \to x^{2} + 2 x - 3 = 0$ $f(x) = 0 -> x^2+2x-3=0$
$\to (x + 3) (x - 1) = 0$ $-> (x+3)(x-1)=0$

Hence $f (x)$ $f(x)$ has zeros at $x = - 3 and 1$ $x=-3 and 1$

$f'(x) = -2x-2$
$f(x)$ has a turning point where $f'(x)=0$

$f'(x) =0 -> -2x-2=0 -> x=-1$

Since the coefficient of $x^2$ is negative $f(-1)$ is a maximum value of $4$

These points can be seen on the graph of f(x) below.

graph{3-x^2-2x [-11.25, 11.25, -5.62, 5.63]}

How do you use the important points to sketch the graph of f(x)=3-x^2-2xf(x)=3−x2−2xf(x)=3-x^2-2x?