Question 134725This question is from textbook College Algebra : I am completely lost on how to do these. Any help would be great. It says: Sketch the graph of the following. Label vertex, x-and y-intercepts. f(x)=x^2-x-12 Vertex=( , ) x-intercepts=( , );( , ) y-intercept=( , ) This question is from textbook College AlgebraFound 3 solutions by cbcrawford978, ankor@dixie-net.com, solver91311:Answer by cbcrawford978(7) (Show Source): Answer by ankor@dixie-net.com(22719) (Show Source): You can put this solution on YOUR website! I am completely lost on how to do these. Any help would be great. It says: Sketch the graph of the following. Label vertex, x-and y-intercepts. f(x) = x^2 – x – 12 Here’s how you can sketch a graph f(x) = y so we have: y = x^2 – x – 12 : We will find the values of y by substituting -5 to +5 in the equation A table of these value will be: x | y ——- -5 |+18 -4 |+8 -3 |0; when y = 0, this is one of the x intercepts -2 |-6 -1 |-10 0 |-12; notice when x = 0, the value of y is 12, this is the y intercept +1 |-12; it has reached a minimum and started back up again +2 |-10 +3 |-6 +4 | 0; this is the 2nd x intercept +5 |+8 Plot these values, it should look like this: : Vertex=( , ) To find the vertex, first find the axis of symmetry. This can be found using the formula x = -b/(2a) In this equation a=1; b=-1 x = x = or .5 Find the vertex by substituting 1/2 for x in the original equation: y = .5^2 – .5 – 12 y= .25 – .5 – 12 y = -12.25 The x/y values for the vertex is: .5, -12.25, this is the minimum as you can see by the graph : x-intercepts=( , );( , x intercepts are the points which the graph crosses the x axis (y=0) you can find them by solving the equation x^2 – x – 12 = 0 Factors to (x-4)(x+3) = 0 x = +4 x = -3; the x intercepts : y-intercept=( , ) As we said the y intercept occurs when x = 0, you see when substitute 0 for x in the equation; y = -12 : Did this help you? Do you have any questions? Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website! Your function is in the form where , , and The vertex is located at the point (,). So, calculate to get the x-coordinate, then evaluate to get the y-coordinate. Plot this point on your graph. The x-intercepts are at (p,0) and (q,0) where and are the solutions to f(x)=0. So solve to determine p and q, then plot these two points. The y-intercept is at (0,f(0)). Evaluate f(0) by substituting 0 for x in . Plot this point. Multiply the x-coordinate of the vertex by 2. Symmetry of the curve about the vertical line means that there will be another point on the curve at (,), i.e. 2 times the x-coordinate of the vertex and the y-coordinate of the y-intercept. Plot this point. Draw a smooth curve through the plotted points. Done. It should look something like this:

You are watching: SOLUTION: I am completely lost on how to do these. Any help would be great. It says: Sketch the graph of the following. Label vertex, x-and y-intercepts. f(x)=x^2-x-12 Vertex=( , ) x-. Info created by GBee English Center selection and synthesis along with other related topics.