A. x intercepts are where the graph hits the x axis or where f(x)=0
0=2x^2-x-10
solve
hmm
we can use the ac method
multiply 2 and -10 to get -20
what 2 numbers muliply to get -20 and add to get -1 (the coefint of the x term)
-5 and 4
split the midd into that
0=2x^2+4x-5x-10
group
0=(2x^2+4x)+(-5x-10)
factor
0=2x(x+2)-5(x+2)
undistribute
0=(x+2)(2x-5)
set each to 0
0=x+2
0=-2
0=2x-5
5=2x
5/2=x
x intercepts are at x=-2 and 5/2 or the points (-2,0) and (5/2,0)
B. ok, so for f(x)=ax^2+bx+c
if a>0, then the parabola opens up and the vertex is a minimum
if a<0 then the parabola opens down and the vertex is a max
f(x)=2x^2-x-10
2>0
opoens up
vertex is minimum
ok, the vertex
the x value of the vertex in f(x)=ax^2+bx+c= is -b/(2a)
the y value of the vertex is f(-b/(2a)) so
given
f(x)=2x^2-x-10
a=2
b=-1
-b/2a=-(-1)/(2*2)=1/4
f(1/4)=2(1/4)^2-(1/4)-10
f(1/4)=2(1/16)-1/4-10
f(1/4)=1/8-1/4-10
f(1/4)=1/8-2/8-80/8
f(1/4)=-81/8
so the vertex is (1/4,-81/8) or (0.25,-10.125)
C. graph the x intercepts and the vertex
the vertex is min and the graph goes through the x intercepts
Answer:
reflection, then rotation, then translation
Step-by-step explanation:
Read the problem statement. It tells you the sequence:
Triangle A B C is <em>reflected</em> across B C to form triangle A prime B C. Triangle A prime B C is <em>then rotated</em> about point A to form A prime B prime C prime. Triangle A prime B prime C prime is <em>then shifted to the right</em> to form A double-prime B double-prime C double-prime.
The transformations are ...
reflection, then rotation, then translation
Thxxxxxxxxxxxxxxxxxxcxxxxxxxxxxxcxxxxx
X = 60°
Because the square in the corner tells you that it is 90° (right angle)
So your equation to solve for x would be...
x + 30 = 90
Now you need to isolate the x.
Subtract both sides by 30.
x + 0 = 90 - 30
x = 60
