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
