To find the maximum height you need to find the vertex:(h,k)
Your equation is in vertex form a(x-h)+k and the vertex is (h,k) where k is the maximum height and the h is the distance it went to reach the maximum height.
k=6 so the kangaroo's maximum height is 6 feet.
To find how long is the kangaroo's jump, take a look at the graph. You will notice that the parabola ends at the distance the kangaroo jumped. You will also see that it is the one of the x-intercepts.
-.03(x-14)^2+6=0
-.03(x-14)^2+6-6=0-6
-.03(x-14)^2=-6
-.03/-.03(x-14)=-6/-.03
(x-14)^2=200
[(x-14)^2]^.5=200^.5
x-14=(200)^.5
x-14+14=(200)^.5+14
x≈28.14 feet
The kangaroo jumped a distance of 28.14 feet.
You will notice that the square root of a number gives you two solutions a positive and a negative one. The other solution is -.14, which we know distance is not negative so we do not use that solution. Also, I used the ^.5 instead of using the square root. It is the same.
Let the width be x.
Length is 8 feet more than width. Length = x + 8
Area = x(x + 8)
width increased by 4, that is, (x + 4)
Length decreased by 5, (x + 8 - 5) = (x + 3)
Area = (x + 4)(x +3)
Area remains the same
x(x + 8) = (x+4)(x +3)
x² + 8x = x(x +3) + 4(x +3)
x² + 8x = x² +3x + 4x +12
x² + 8x = x² +7x +12 Eliminate x² from both sides
8x = 7x + 12
8x - 7x = 12
x = 12
Dimensions of original rectangle : x, x + 8
12, 12 +8 = 12, 20
Original rectangle is 20 feet by 12 feet
Answer:
y=2x-2
Step-by-step explanation:
<em>the slope is y/x</em>
the slope is 2 or 2/1
<em>subtract 2 from the y value and 1 from the x value</em>
(3-1,4-2) = (2,2)
<em>keep doing this until you get a 0 in the x value</em>
(2-1,2-2) = (1,0)
<em>1 is your x-intercept</em>
(1-1,0-2) = (0,-2)
-2 is your y intercept.
So now you know your y-intercept and your slope so you can now write your equation.
<em>y=mx+b</em>
<em>m=slope, b=y-intercept</em>
m=2, b=-2
<em>substitute into the equation</em>
y=2x-2
One of the ways to graph this is to use plug in a few x-values and get an idea of the shape. Since the x values keep getting squared, there is an exponential increase on either side of the y-axis. You can see this by plugging in a few values:
When
x=0,f(x)=0
x=1,f(x)=1^2=1
x=2,f(x)=2^2=4
x=3,f(x)=3^2=9
x=4,f(x)=4^2=16
The same holds true for negative x-values to the left of the y-axis since a negative value squared is positive. For example,
x=−1,f(x)=(−1)2=1*−1=1
x=2,f(x)=(−2)2=−2*−2=4
The graph of f(x)=x^2 is called a "Parabola." It looks like this:
It is A
Hope this helps :3