About Chanda Walker Expertise I can answer word problems involving mathematics at the high school and college level. I particularly enjoy calculus word problems. Please don't just type the math problem without comments. If you don't tell me what problem your having, I can't help.
Experience Have done word problems as a tutor and as a student of mathematics and physics for years.
Organizations Sigma Xi
Education/Credentials Optical Science PhD
Past/Present Clients I've answered hundreds of questions here at AllExperts.com in algebra, physics and general math sections.
Question Okay, thank you for looking. I am having serious problems with my grade 10 math. If you could go into some detail about how to approach and complete these questions, it would be GREATLY appreciated! Okay so here are the questions:
13. A golf ball is hit and its height is given by h = -4.9t^2 + 29.4t, where h is its height in meters and t is the time in seconds.
a) At what time does the golf ball reach its maximum height?
b) What is the ball’s maximum height?
14. Without drawing a graph, determine if y = (x + 2)2 + 4 has any zeroes. Explain.
15. A parabola passes through the points (0, 0), (4, 0), and (5, -5), without graphing determine the equation of this parabola.
And how do you graph y = 2x^2 + 4x – 6 by finding the zeroes and the vertex?
Thank you SO much!
Leah :)
Answer Whoa! No killing quadratics. That is an insult to a mathematician!
To find a maximum of a quadratic, all you have to do is apply this formula:
tmax = -b/2a
You'll be able to derive that in calculus but for now just memorize it.
In your first equation, a = -4.9 and b = 29.4.
t = 29.4/(9.8)
= 3 seconds
To find the maximum height AFTER you found the time of the maximum, just plug your answer for t into the equation.
For your equation can (x + 2)^2 equal a negative 4? No, it can't so it doesn't have any zeros.
y = ax^2 + bx + c
Since (0,0) is a point, c must be zero. Just substitute 0 for y and x and you'll solve that.
Since we know c, we can solve for the other variables now.
0 = 16a + 4b
and
-5 = 25a + 5b
Algebra will solve for a and b from here. b = -4a comes from the first equation. Then
-5 = 25a - 20a
-5 = 5a
a = -1
b = 4
After you find the zeroes and the vertex, all you have to do is draw a reasonable curve connecting the first zero to the vertex and then the vertex to the second zero. Continue the lines past the x-axis.
I don't do graphing. Sorry.
The zeroes are found by:
y = 2(x^2 + 2x - 3)
y = 2(x + 3)(x - 1)
x = -3
x = 1
The vertex is between the zeros at x = -1 and the y value is found by substitution:
