How do you graph $f (x) = \frac{2}{3} x^{2} - 5$ $f(x) = 2/3x^2 - 5$ ?

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Answer 1 · 2017-08-03T18:38:18

See a solution process below:

Setting $x$ $x$ to $0$ $0$ gives us the point:

$f (x) = (\frac{2}{3} \cdot 0^{2}) - 5$ $f(x) = (2/3 * 0^2) - 5$

$f (x) = - 5$ $f(x) = -5$

Or

$0, - 5$ $0, -5$

We can find the $0$ $0$ s by solving the function for $0$ $0$ :

$\frac{2}{3} x^{2} - 5 = 0$ $2/3x^2 - 5 = 0$

$\frac{2}{3} x^{2} - 5 + 5 = 0 + 5$ $2/3x^2 - 5 + 5 = 0 + 5$

$\frac{2}{3} x^{2} - 0 = 5$ $2/3x^2 - 0 = 5$

$\frac{3}{2} \times \frac{2}{3} x^{2} = \frac{3}{2} \times 5$ $3/2 xx 2/3x^2 = 3/2 xx 5$

$x^{2} = \frac{15}{2}$ $x^2 = 15/2$

$x^{2} = \pm \sqrt{\frac{15}{2}}$ $x^2 = +-sqrt(15/2)$

Or

$(- \sqrt{\frac{15}{2}}, 0)$ $(-sqrt(15/2), 0)$ and $(- \sqrt{\frac{15}{2}}, 0)$ $(-sqrt(15/2), 0)$

graph{(y-((2x^2)/3)+5)(x^2+(y+5)^2-0.125)((x+(15/2)^(1/2))^2+(y)^2-0.125)((x-(15/2)^(1/2))^2+(y)^2-0.125)=0 [-20, 20, -10, 10]}

How do you graph f(x) = 2/3x^2 - 5f(x)=23x2−5 f(x) = 2/3x^2 - 5?