Hilda adds 5 to a number, then multiples the sum by -2. The result is 6. Write and equation to find the number, x. What is the n
1 answer:
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OK. All the work is in the attached drawing.
Please spend enough time with it to understand what was done and how, so that you can do the next problem like this one completely on your own.
Good luck !
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Here's a picky pickypoint to think about concerning the "cost". I'm not quite sure what to do about this:
The initial, one-time $55 is a "deposit". If she doesn't smash the bike or steal it, she gets the $55 back when she returns the bike.
So if she eventually gets the $55 back, <u><em>is it a cost</em></u> ? ? I don't know how to think about it.
She does need to <em>give</em> them the $55 at the beginning, just to take the bike out. So she has to have it with her and give it to them temporarily, even though she'll get it back, and she'll still have it when she returns home.
So here's the story:
-- While she HAS the bike, the graph correctly shows all the money she needs, in order to get the bike, and keep it for however long she keeps it.
-- Finally, at the end of the week, after she returns the bike and gets her $55 back, the line and everything on the graph will shift down by $55. The line will start at zero, and the little red ordered pairs will also shift straight down and still be on the line.
To put it one more way:
-- While Jen has the bike, the y-intercept of the graph is $55.
-- After she returns the bike in good condition, the y-intercept is zero.
Answer:
A, D, E are true
Step-by-step explanation:
You have to complete the square to prove A. Do this by first setting the function equal to 0, then moving the 5 to the other side.
Now we can complete the square. Take half the linear term, square it, and add it to both sides. Our linear term is 8 (from the -8x). Half of 8 is 4, and 4 squared is 16. So we add 16 to both sides.
We will do the addition on the right, no big deal. On the left, however, what we have done in the process of completing the square is to create a perfect square binomial, which gives us the h coordinate of the vertex. We will rewrite with that perfect square on the left and the addition done on the right,
Now we will move the 11 back over, which gives us the k coordinate of the vertex.
From this you can see that A is correct.
Also we can see that the vertex of this parabola is (4, -11), which is why B is NOT correct.
The axis of symmetry is also found in the h value. This is, by definition, a positive x-squared parabola (opens upwards), so its axis of symmetry will be an "x = " equation. In the case of this type of parabola, that "x = " will always be equal to the h value. So the axis of symmetry is
x = 4, which is why C is NOT correct, either.
We can find the y-intercept of the function by going back to the standard form of the parabola (NOT the vertex form we found by completing the square) and sub in a 0 for x. When we do that, and then solve for y, we find that when x = 0, y = 5. So the y-intercept is (0, 5).
From this you can see that D is also correct.
To determine if the parabola has real solutions (meaning it will go through the x-axis twice), you can plug it into the quadratic formula to find these values of x. I just plugged the formula into my graphing calculator and graphed it to see that it did, indeed, go through the x-axis twice. Just so you know, the values of x where the function go through are (.6833752, 0) and (7.3166248, 0). That's why you need the quadratic formula to find these values.