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the coefficient of x will be negative for a parabola that opens downwards the function -x^2 is this shape with the vertex at the origin (0,0) But we want to move this up to (0,36) by adding 36 so our equation would be -x^2 + 36
Thanks! what about this part? Create a table of values for a linear function. A drone is in the distance, flying upward in a straight line. It intersects the rainbow at two points. Choose the points where your drone intersects the parabola and create a table of at least four values for the function. Remember to include the two points of intersection in your table.
well we can choose one value that's on the y-axis say (0,20) for example. Then one value on left side of our rainbow say x = -5 and y will then be -(-5^2) + 36 = 11
we can then work out the equation of the line and find the point where it intersects another point on the curve and another point before or after the intersections
Don't we need to create a table w/ 2 intersection points too?
slope of line is (20-11)/ 0-(-5) = 9/5
yes we have one already and we can find the other by solving the 2 equations. the line cuts the axis at y=20 so its equation is y = (9/5) x + 20
How would we do that? Just plug in points?
so another point could be at x = -7 and y= (9/5)*-7 + 20
to find the second point of intersection we need to solv (9/5)x + 20 = -x^2 + 36
would we subtract 20 from both sides?
its a quadratic so add x^2 to both sides and subtract 36 from both sides
x^2 + (9/5)x - 16 = 0
now what do we do?
use the quadratic formula
Ok! But wouldn't the solution be complex making it no real solution?
no one solution will be negative though which you can ignore
the solution will be close to 3 so that will give you your 4 points
alrighty thank you!