# How do you sketch the graph f(x)=x^3+1?

Jul 5, 2016

see explanation

#### Explanation:

write as:$\text{ } y = {x}^{3} + 1$

At x=0; y=1 larr " y-intercept"
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$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} \implies \text{local max/min at } x = 0 \to \left(0 , 1\right)$

Only one $x$ value for local max/min so the graph gradient goes from positive to 0 to positive again (no negative gradient)
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At $y = 0 \implies {x}^{3} = - 1 \leftarrow \text{ x-intercept}$
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For 2 more points set $x = \pm 2$

At $x = - 2 : {x}^{3} + 1 = - 7 \to \left(- 2 , - 7\right)$

At $x = 2 : {x}^{3} + 1 = + 9 \to \left(+ 2 , + 9\right)$