The graph of f(x)=x2−2x−3 is shown above. a) Determine the x-intercepts of the graph. (Note: List the value(s) of x separated by a comma.) b) Find the solution(s) for: x2−2x−3=0

Question

The graph opened upwards with a vertex on (1,-4). Not sure if this is needed but included it since the graph image was too large to include with this description.

Answer 1

x^2 -2x -3

= (x-3)(x+1)

set each factor = 0 and solve for x

x=3, -1 are the two x intercepts or the points (3, 0), (-1, 0)

y intercept is y= -3 or the point (0, -3)

it's an upward opening parabola

complete the square to find the vertex

x^2 -2x -3

x^2 -2x + 1 -3-1

(x-1)^2 -4 in vertex form vertex = (1,-4)

solutions of x^2-2x-3 =0 are the x intercepts, or roots x=3 or -1

3^2 -2(3) -3 = 9-6-3 = 0

(-1)^2 -2(-1) -3 = 1+2-3=0

using Descartes' Rule of Signs there should be maximum one potential positive solution

signs change one time +x^2 to -2x

then for f(-x) signs change just one time, so maximum one potential negative solution

turns out, exactly one of each, 3 and -1

f(-x) = (-x)^2 -2(-x) -3 = x^2 +x -3. sign changes once, from +x to -3

it's upward opening due to the positive x^2 term and the vertex is (1,-4) = minimum point, so it has to cross the x axis twice. Minimum point is below the x axis, and it's upward opening. the graph has to intersect the x axis.

had the vertex been above the x axis, there would have been no x intercepts and the solutions to the polynomial set = 0 would have been imaginary, two imaginary solutions