Of 20.408 m, then h decreases again to zero, as expected. `t = -b/(2a) = -20/(2 xx (-4.9)) = 2.041 s `īy observing the function of h, we see that as t increases, h first increases to a maximum What is the maximum value of h? We use the formula for maximum (or minimum) of a quadratic function. It goes up to a certain height and then falls back down.) (This makes sense if you think about throwing a ball upwards. We can see from the function expression that it is a parabola with its vertex facing up. So we need to calculate when it is going to hit the ground. The domain is all x-values or inputs of a function and the range is all y-values or outputs of a function. Also, we need to assume the projectile hits the ground and then stops - it does not go underground. The range is all the values of the graph from down to up. Generally, negative values of time do not have any Have a look at the graph (which we draw anyway to check we are on the right track): So we can conclude the range is `(-oo,0]uu(oo,0)`. We have `f(-2) = 0/(-5) = 0.`īetween `x=-2` and `x=3`, `(x^2-9)` gets closer to `0`, so `f(x)` will go to `-oo` as it gets near `x=3`.įor `x>3`, when `x` is just bigger than `3`, the value of the bottom is just over `0`, so `f(x)` will be a very large positive number.įor very large `x`, the top is large, but the bottom will be much larger, so overall, the function value will be very small. As `x` increases value from `-2`, the top will also increase (out to infinity in both cases).ĭenominator: We break this up into four portions: To work out the range, we consider top and bottom of the fraction separately. So the domain for this case is `x >= -2, x != 3`, which we can write as `[-2,3)uu(3,oo)`. (Usually we have to avoid 0 on the bottom of a fraction, or negative values under the square root sign). In general, we determine the domain of each function by looking for those values of the independent variable (usually x) which we are allowed to use. For a more advanced discussion, see also How to draw y^2 = x − 2.
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