Help with the Quadratic Equation! Will give medals for correct answer!
Where is the question : ^)
A ball is launched from a sling shot. Its height, h(x), can be represented by a quadratic function in terms of time, x, in seconds. After 1 second, the ball is 121 feet in the air; after 2 seconds, it is 224 feet in the air. Find the height, in feet, of the ball after 3 seconds in the air.
I don't understand how to do these. I don't understand the lesson and my teacher just tells me to look at the lesson. If you could explain how to do it, that'd be great, too
I got 327 for the height but I don't know.
I probably did it wrong : ^)
Yeah, it was incorrect :^(
r.i.p
My apologies.
It's okay.
Any quadratic equation can be put in the form \[y = ax^2 + bx + c\]where \(a,b,c\) are constants and \(a\ne0\) We have two points, (\(1,121\)) and (\(2,224\)), where the x value is the time in seconds (the independent variable) and the y value is the height (the dependent variable). We can also assume a third point: (\(0,0\)) representing the state before the ball is launched Plug all the points into the equation and you have 3 equations in 3 unknowns, which you can then solve to find the values of \(a,b,c\). \((0,0)\): \[0 = a(0)^2 + b(0) + c\] \((1,121)\): \[121 = a(1)^2 + b(1) + c\] \((2,224)\): \[224 = a(2)^2 + b(2) + c\] I think you will find the value of \(c\) quite easy to determine :-) When you have found \(a\) and \(b\), use them (along with \(c\)) to fill out your quadratic formula, then evaluate it at \(x=3\) for the answer. You should find that your answer starts with a 3.
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