How do you graph #y=x^2-2x-3#?

Question

Genesis Allen · Accepted Answer

Find and plot: the vertex, the #x# intercepts, the #y# intercept (and any other test values for #x# you like); then connect the dots.

#y=x^2-2x-3 =(x-3)(x+1)# #y# intercept #= -3##color(white)("XXXX")# (found by setting #x=0#) #x# intercepts #=3# and #=-1##color(white)("XXXX")#(from the factors) vertex: #(x,y) = (1,-4)##color(white)("XXXX")#(#x# by completing the square, #y# by substitution) graph{x^2-2x-3 [-5.3, 7.19, -4.805, 1.435]}

Jameson Allen · Answer

undefined To graph the equation y = x^2 - 2x - 3, you can follow these steps:

1. Plot the y-intercept by substituting x = 0 into the equation to find the y-coordinate. In this case, y = 0^2 - 2(0) - 3 = -3. So, the y-intercept is at the point (0, -3).

2. Find the x-coordinate of the vertex by using the formula x = -b/(2a), where a = 1 (coefficient of x^2) and b = -2 (coefficient of x). So, x = -(-2)/(2*1) = 1. This gives the x-coordinate of the vertex.

3. Find the y-coordinate of the vertex by substituting the x-coordinate found in step 2 into the equation. y = (1)^2 - 2(1) - 3 = -4. So, the vertex is at the point (1, -4).

4. Plot additional points on either side of the vertex to sketch the parabola. You can choose other x-values and substitute them into the equation to find corresponding y-values.

5. Draw a smooth curve through the plotted points to represent the graph of y = x^2 - 2x - 3.

6. Label the axis and any key points such as the vertex and intercepts to complete the graph.