The graph gives no reason to expect that y=0 is excluded, so we conclude the range is ...
... y ≥ 0
Answer:
a=8
c=5.5
Step-by-step explanation:
solve simultaneously:
2a + 4c = 38.00
3a + 3c = 40.50
times the first eq. by 3 and the second by 2
6a + 12c = 114
6a +6c = 81
subtract from one another
6a + 12c = 114
-6a +6c = 81
6c = 33
c=5.5
3a + 3x5.5 = 40.50
3a= 24
a=8
you can subtract from point 1 to point 2
(9,-2) - (8,3)
9-8= 1
-2-3= -5
therefore horizontal leg is 1
vertical leg is -5
you can also sketch and count the units :)
Question: Simplify
.
First off, you <em>never</em> want square roots on the bottom of a fraction, so let's multiply both sides by
to get rid of it. All we have to do from there is simplify!
Remember that the vertex form of a parabola or quadratic equation is:
y=a(x-h)^2+k, where (h,k) is the "vertex" which is the maximum or minimum point of the parabola (and a is half the acceleration of the of the function, but that is maybe too much :P)
In this case we are given that the vertex is (1,1) so we have:
y=a(x-1)^2+1, and then we are told that there is a point (0,-3) so we can say:
-3=a(0-1)^2+1
-3=a+1
-4=a so our complete equation in vertex form is:
y=-4(x-1)^2+1
Now you wish to know where the x-intercepts are. x-intercepts are when the graph touches the x-axis, ie, when y=0 so
0=-4(x-1)^2+1 add 4(x-1)^2 to both sides
4(x-1)^2=1 divide both sides by 4
(x-1)^2=1/4 take the square root of both sides
x-1=±√(1/4) which is equal to
x-1=±1/2 add 1 to both sides
x=1±1/2
So x=0.5 and 1.5, thus the x-intercept points are:
(0.5, 0) and (1.5, 0) or if you like fractions:
(1/2, 0) and (3/2, 0) :P
