Find the x-intercepts for the parabola defined by the equation below. y = x2 + x - 12
The parabola intercepts the x-axis happen when? Well, it's when "y" = 0. So you want to solve for x, 0 = 2x^2 + 2x -4 0 = 2x^2 + 2x -4 = (2x -2)(x + 2). So 2x -2 = 0 OR x+2=0 are the two solutions. So in each case, solve for x, and those will be your answers. Ok. So by definition the x-intercept of a line/graph is when y=0 (since at that point will then be "on the x axis"). for example (1,0), (2,0), -3,0) etc... are all points that intercept the x-axis - as you can see if you graph them. That makes perfect sense, since the y-coordinate is 0. So given the graph (in this case a parabola), will intercept the x-axis when the y coordinate is 0, go ahead and replace "y" with "0" in the initial equation and solve for x 0 = 2x^2 + 2x -4 when you see something like this (called a quadratic equation because it's in the form ax^2+bx+c =0), there are two ways to solve it once is to use the quadratic formula - because in this format you know the values of a, b, and c http://en.wikipedia.org/wiki/Quadratic_equation (which in this case are 2, 2, and -4) The second way is to factor the equation. That is change it into 2 terms that are equivalent when multiplied together. 2x^2 + 2x -4 = (2x -2)(x + 2) Now once it is factored, you'll notice something interesting about the right hand side *anything* multiplied by zero is also zero - which is exactly what you want y to be. and hence just by factoring, you can also get both solutions. when *either* 2x -2 = 0, OR x+2=0, you know the whole thing will be zero too (since something gets multipled by 0) And so 2x-2 =0, so in that case x= 1. in the second case x+2 =0 , so x=-2. So you know that the parabola intercepts the x axis when x=1, or x=-2, so at (1,0) and (-2,0)
wait so whats the answer out of these 4 answers? (-12, 0) and (1, 0) (-4, 0) and (3, 0) (0, -4) and (0, 3) (0, -12) and (0, 1)
@Muzzack ^
D
thank you so much
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