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
To find an average you take the numbers to be averaged and add them together then You take the number and divide it by the number of numbers, in this case 4, and that gives you the average.
on this problem you want to do the opposite. you know the average she wants, an 87, so you multiply it by 4. 87*4=348. then you add the three numbers you have (88,81 and 87) and you get 256.
then you subtract 256 from 348 and you get 92. 92 is your answer to check just add 87+81+88+92=348
348/4=87
87 was the average she wanted.
Hope this helped! good luck :)
Absolutely not because
if you see as fraction
8/7 and 15/16 is not equal at all
if it had to be equal if should've been 16/14
