SOLUTION: Hello i need help with this question thank you so much for your help and please show the answers as clear as possible. A clay pigeon is fired into the air. It moves along a pa

Question 1180273: Hello i need help with this question thank you so much for your help and please show the answers as clear as possible.

A clay pigeon is fired into the air. It moves along a path according to the equation h(x)=-5x^2++14x. h(x) represents the height the pigeon travels above ground, and x represents the horizontal distance. A person fires a bullet which goes in a straight line according to the formula b(x)=0.5x+3.25.x is still the horizontal distance, along the ground. Distances are in metres.
It is possible that the bullet can hit the clay pigeon in two places along h(x), depending on when the person fired the rifle. What are those two places (answer in terms of horizontal distance and height above ground)? All numbers should be accurate to the nearest tenth.
A helpful formula is the quadratic formula -b±√b^2-4ac/2a

for quadratics of the form: ax2+bx+c=0.

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Can graph this just to see what it looks like Note that I am assuming b(x)=0.5x+3.25. I am ignoring the .x" because the formula as I wrote it is a linear equation.

Intersect when the two are equal
-
so -5x^2+14x=0.5x+3.25
0=5x^2-13.5x+3.25. I like dealing with positive exponents in the square.
Using the quadratic formula where a=5, b=-13.5 and c=3.25
x=(1/10)(13.5+/- sqrt (117.25)); sqrt term=10.83
x=0.267 feet and 2.433 sec or 0.3 sec and 2.4 sec
-
h(.267)=3.38 feet substituting .267 into both equations (1 is sufficient, but 2 is a check)
h(2.433)=4.47 feet. Don't round until the last step. Keep all decimal places in x as you calculate. It's much easier to substitute into b(x), since there is no square, but one should check both, and make sure it makes sense on the graph.
The horizontal lines on the graph are at 3.38 and 4.47 feet. They should intersect the graph where the straight line and the curve meet, and they do.

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