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posledela
3 years ago
12

How to reflect over y axis in an equation?

Mathematics
1 answer:
iren [92.7K]3 years ago
7 0
<span>1)  y = -f(x) (This is the reflection about the x-axis of the graph y = f(x).)  That is for every point (x, y) there is a point (x, -y).
</span><span>2)  y = |f(x)| means that the entire graph will be above the x-axis.  Why?  (The absolute value is always positive, that's why!!)<span>  To graph the absolute value graph, graph the function y = f(x).  Anything above the x-axis, stays above it, anything below the x-axis is reflected above the x-axis and anything on the x-axis, stays on the x-axis.
</span></span><span>3)  y = f(-x)  (This is reflection about the y-axis of the graph y = f(x))  For every point on the right of the y-axis, there is a point equidistant to the left of the y-axis.  That is for every point (x, y), there is a point (-x, y).  
</span><span>4)  Reflections about the line y = x is accomplished by interchanging the x and the y-values.  Thus for y = f(x) the reflection about the line y = x is accomplished by x = f(y).  Thus the reflection about the line y = x for y = x2 is the equation x = y2. </span>
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Answer:

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Step-by-step explanation:

Using y - y₀ = vt - 1/2gt² where g = 32 ft/s², and v the velocity of the football

So y = y₀ + vt - 1/2 × (32 ft/s²)t²

y = y₀ + vt - 16t² where y₀ = 6.5 ft

y = 6 + vt - 16t²

Now, when t = 3.5 s, that is the time the teammate catches the ball after the quarterback throws it, y = 5 ft. Substituting these into the equation, we have

5 = 6.5 + v(3.5 s) - 16(3.5 s)²

5 = 6.5 + 3.5v - 196

collecting like terms, we have

5 - 6.5 + 196 = 3.5v

194.5 = 3.5v

v = 194.5/3.5 = 55.57 ft/s ≅ 55.6 ft/s

So, substituting v into y, our quadratic model is

y = 6 + 55.6t - 16t²

re-arranging, we have

y = - 16t² + 55.6t + 6

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