Question #119766
Given the function f(x) =2xsquared - x-1
(a) Is the point (-1, 2) on the graph of f .
(b) If x=-2, what is f(x) ? What point is on the graph of f ?
(c) If f (x)=-1, what is x ? What point(s) are on the graph of f ?
(d) What is the domain of f ?
(e) List the x - intercepts, if any, of the graph of f .
1
Expert's answer
2020-06-07T15:59:22-0400


(a) Yes, the point (-1, 2) on the graph of f because f(1)=2(1)2(1)1=2+11=2.f(−1)=2∗(−1)^2−(−1)−1=2+1-1=2.

(b) If x=2x=−2 then f(2)=2(2)2(2)1=24+21=9.f(−2)=2∗(−2)^2−(−2)−1=2*4+2-1=9.

Point (-2, 9) is on the graph of f.

(c) If f(x)=1f(x)=−1 then 2x2x1=12x^2−x−1=−1, 2x2x=02x^2-x=0, x(2x1)=0,x(2x−1)=0, x=0x=0 or x=0.5x=0.5.

Points (0, -1) and (0.5, -1) are on the graph of f.

(d) f is a quadratic function, its domain is all real numbers.

(e) In order to find x-intercepts of the graph of f we take y=0 and solve the quadratic equation: 2x2x1=02x^2−x−1=0 .

a=2,b=1,c=1a=2, b=−1, c=−1 .

x=1±(1)242(1)22.x=\frac{1\pm \sqrt{(-1)^2-4*2*(-1)}}{2*2}.

x=1±34x=\frac{1\pm 3}{4} .

x=0.5x=−0.5 or x=1.x=1.

So we have two x-intercepts of the graph of f: (-0.5, 0) and (1, 0).


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