That said, let's use the general form again and set the result of the function to 0 and try to solve for x. There can be 0 or 2 x-intercepts depending on the value of k and a.Īnd no x-intercepts otherwise. The method I thought of to find the x-intercepts a bit more involved, maybe someone else knows an easier way. Taking one of the examples, f(x)= |x−1| + 5 where a=1, h=1 and k=5: the y-coordinate of the intercept is 1|1| + 5 = 6, which means the intercept is at (0,6). Which gives us, as a general rule, (0,(a|h|+k)) as the y-intercept. To find the y-intercept, we can set x to 0. As for the y-coordinate: since we just saw that |x-h| = 0, a|x−h| must also be 0, which only leaves us with k. This is the same as saying x = h, which gives us the x-coordinate of the vertex. The minimum or maximum (depending on whether a is positive or negative) of the graph is at the point where x - h = 0. Just to recapitulate, the general form is: ![]() Sorry if it's too much of a wall of text to get through. I'm not an expert here, but it was an interesting exercise to figure out the answer to your questions and I figured I might as well post it here. Let me know if that didn't help, or if there is a specific function you are struggling with, or maybe would even like some to try out. Also, keep in mind that even if inside the absolute value bars if b was negative, outside it becomes positive. The point being you always want x by itself for this. If you ever have something like a|bx-h|+k where there is a number in front of the x you need to get rid of it if you are not aware of factoring this is what it would look like a|b||x - h/b|+k where a|b| becomes the new "a" and h/b becomes the new h, then you would solve it normally. then, since it's an absolute value function you need to know that the same line goesalong the left to make that V shape, so -5 would mean on the left down 3 and left 1. so if a was -3 that's down 3 right 1 using rise over run. treat it like a linear equation where a is the slope. so if it helps, the x coordinate is kinda backwards.Īfter the V tip you then look at a. if you wonder why it is -5 even though we are adding 5, you just need to look at the original a|x-h|+k if we had -5 then it would be just like that, but since it is +5, we have to look at it as -5, minus negative 5. ![]() specifically it moves the tip to (h,k) so if you have |x+5|-7 then the tip of the V shape goes to (-5,-7). ![]() Normally the tip of the V shape is at (0,0) this changes depending on h and k. a|x-h|+k Specifically you want to look at h and k first. When you have an absolute value function you want to look at what are in the places of a, h and k.
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