How do you graph $y = (- \frac{3}{4}) x^{2}$ $y = (-3/4) x^2$ ?

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How do you graph $y = (- \frac{3}{4}) x^{2}$ $y = (-3/4) x^2$ ?

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Answer 1 · 2017-04-14T14:51:36

See explanation

Explanation:

The coefficient of $x^{2}$ $x^2$ is negative so the general shape of the graph is $\cap$ $nn$

As the coefficient is less than 1 it has the effect of widening the 'base graph' of $y = x^{2}$ $y=x^2$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Write as: $y = - \frac{3}{4} x^{2} + 0 x + 0$ $y=-3/4x^2+0x+0$

Compare to the standardised equivalent of:

$y = a x^{2} + b x + c$ $y=ax^2+bx+c$

The y-intercept $\to c = 0$ $->c=0$

The x-intercept is at $y = 0$ $y=0$ giving $0 = - \frac{3}{4} x^{2} \Rightarrow x = 0$ $0=-3/4x^2 => x=0$

As there is only 1 solution ( $x = 0$ $x=0$ ) the x-axis is tangential to the vertex. Also, as this is at $y = 0$ $y=0$ we have:

Vertex $\to (x, y) = (0.0)$ $->(x,y)=(0.0)$

Other than this and you will to build a table of values. Plot the points and draw the best plot you can that passes through those points.

Tony B

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How do you graph $y = (- \frac{3}{4}) x^{2}$ $y = (-3/4) x^2$ ?

1 Answer

Explanation:

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