In Exercises 1–4, use the vertex and intercepts to sketch the ...

Question

In Exercises 1–4, use the vertex and intercepts to sketch the graph of each quadratic function. Give the equation for the parabola's axis of symmetry. Use the graph to determine the function's domain and range.

$f (x) = - (x + 1)^{2} + 4$

Accepted Answer

Hello everyone in this video. We're going to be looking at this practice problem where we want to graph the proble defined by F of x is equal to negative X squared minus eight X minus 12. We want to graph it by using its vertex and intercepts and once we have it graft we want to identify the axis of cemetery and the domain and range. So the equation that were given for the problem is f of x is equal to negative X squared minus eight X minus 12. And that form follows the standard form of the quadratic equation which is F of X is equal to a X squared plus bx plus C. And this can help us because the quadratic equation in standard form we can get the for a text value will say the vertex is equal to H. K. And we can get H K. Because H is equal to negative B divided by two A and K is equal to F of H. So if we plug in H back into the equation so we can use this to help us find our vertex and our intercepts. So recall that for a problem, I draw an example here problem that follows this form. The vertex would be where it changes direction going from decreasing to increasing. And in this case the intercept the y intercept is where it crosses the y axis, which would also be the same point here. And then we have the X intercepts where the problem crosses the X axis. So these are all the points that we're going to find to help us graph our problem. But first we need to figure out what A B and C are in our equation. So in this case A. Is equal to negative one, C. There's a negative in front of the X squared B. is equal to -8 and C is equal to -12. So from this we can determine H which is equal to negative negative B. Which is negative eight divided by two A. Which is negative one. So myself for this I have negative times negative eight is positive eight divided by two times negative one is negative two. So H is equal to negative four. Now we need to plug in -4 or X in our equation. So I'll have F of H which is equal to f of negative four, which equals negative negative four squared -8 times negative four -12. I solved that. I have negative four squared is positive 16 but I still have this negative on the outside. So this becomes negative -8 times negative four is positive minus 12. So negative 16 plus 32 is Pososos 16 -12. It's equal to four which means that K. is equal to four because K. is equal to F of H. So if I write the coordinates out for the vertex it would be hK which is negative for positive for and I'm going to go ahead and box that so we don't lose track of it. And now we can solve for the why intercept. And as we mentioned before, the y intercept is where it crosses the y axis, which means that x is equal to zero this point here. So if I make X equal to zero for our equation we have Y is equal to negative zero squared minus eight times zero minus 12. If I simplify that, I have zero minus zero minus 12 or Y is equal to negative 12, which means that our y intercept is at zero negative and I'll go ahead and box that so we don't lose track of it either. And we can also solve the X intercept. And unlike the y intercept now Y is equal to zero for the X intercept, which means our equation will look like zero is equal to negative x squared minus eight, X minus 12. Well in this case because we're trying to find the value of X in the quadratic in a quadratic equation, we're better off using the quadratic formula which is X is equal to negative B plus or minus the square root of B squared minus four A C divided by two. A. And recall that we defined a equal to negative one, B equal to negative eight and C equal to negative 12. So we can use the quadratic equation to help us find the value of X. So if we plug those in, we have negative negative eight plus or minus the square root of negative eight squared minus four times a which is negative one times C which is negative 12 Divided by two times a which is -1. So if I simplify this top row I have eight plus or minus. The square root of negative eight squared is positive 64. Negative four times negative one is positive four, So positive four times negative 12 Divided by two times -1 is -2. And if I continue to simplify I have eight Plus or minus the square root of 64 minus four times 12 which is negative Divided by -2. So finally this Radical simplifies to eight plus or minus the square root of 64 -48 which is 16 divided by negative two and the square root of is plus or -4. So it stays eight plus or minus four divided by two. And because we have this plus or minus we know that there's two sort of scenarios or the x intercept we could have eight plus four divided by two or we could have eight minus four divided by two. So we have eight plus four divided by two. We would have 12 divided by two. Draw a line to separate the two and 12 divided by two is 6. So X equals six which means that one of the x intercepts is at 60. But if we have eight minus four divided by 28 minus four is equal to four divided by two, Which means that X is equal to two. So we have another X intercept too zero and I'll go ahead and box them so that we don't lose them. So now that we have these preliminary values, we can go ahead and look at the graphs and see which ones follow these points. So in this case we're looking for a vertex at -4. Or and you can see that option C has the incorrect vertex because it's a positive for positive force. So I'm gonna go ahead and eliminate option C. And if we look at D it also has the incorrect vertex so we can eliminate it Because we need a negative four as the X. Value and A. And B both have a vertex at negative four four. So that's good. But B has the Y intercept at 0 20 instead of zero, negative 12. So B is eliminated. And that leaves A. As our only answer. But let's go ahead and double check to make sure that it also has the X intercepts. So it needs an X intercept at 60 and 20. Well in this case it has the x intercepts at -60 and -20. But if we go back to the work you can see that I forgot to transfer this negative over to the following equation. So it would be X is equal to negative six and X is equal to negative two. So in fact A is the correct answer. And you can see from the graph that we can plug in any X values, which means that the domain is all real numbers. And the range would be the Y values. And you can see this proble has a max of four, which means that we can only have four and below for the Y values. So the range would be from negative infinity two positive for with a bracket because that's where our vertex is at. So A becomes our final answer. Hopefully this video helped and see you next time.

In Exercises 1–4, use the vertex and intercepts to sketch the graph of each quadratic function. Give the equation for the parabola's axis of symmetry. Use the graph to determine the function's domain and range. $f (x) = - x^{2} + 2 x + 3$

Key Concepts

Vertex of a Quadratic Function

Axis of Symmetry

Domain and Range of Quadratic Functions

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