@phi Can you please help me with this test?!? I have only need help with one thing.
Can you please help me get the h(x) graph into a equation?
Do you know the "vertex form" of the equation for a parabola?
y=a(x-h)^2+k (h,k) is the vertex
what is the vertex for your parabola ?
1,3
are you sure the the vertex is at 1,3 ?
I think so I really can't see, thats part of my problem the graphs are too small.
Here is where the point (1,3) is on your graph
oh duh.... so -1,-3
(although I am sure the vertex is NOT -1,-3)
You fixed the y value and made the x value wrong.
Here is where (-1,-3) is
1,-3
Now (finally!) Replace h and k in your equation with those numbers. after you do that, what do you have ?
h(x)=a (x-1)^2-3
next step is to find "a" to do that, find a "nice" spot on the graph where you can read off the x and y values. Can you find a point where the graph goes exactly through where the lines cross, and then read the what point that is ?
0,-1
now replace h(x) with -1 and x with 0 in your equation. what do you get ?
-1=(0-1)^2-3 ?
what happened to the "a" ?
so its h(x)=-1(0-1)^2-3
No. you start with h(x)=a (x-1)^2-3 which by the way is the same as y = a (x-1)^2 -3 replace x with 0 and y with -1 (because (0,-1) is on the parabola) you leave the "a" alone. try again
-1=a(0-1)^2-3
can you simplify that?
remember PEDMAS parentheses, exponents, divide/multiply, add/subtract is the order of operations.
inside the parens you have (0-1) what does that simplify to ?
2
are you saying that 0 - 1 is 2 ?
simplifying the whole thing is 2 i think
-1= a(0-1)^2 - 3 0-1 is -1 \[ -1 = a(-1)^2 - 3 \] -1*-1 is 1 and a*1 is a \[ -1 = a - 3 \] add +3 to both sides \[ 3-1= a -3 + 3 \\2=a\]
that is the last missing piece in your equation.
so h(x)=2(x-1)^2-3
yes, that looks good.
Thank you!!
yw
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