in-exercises-59-64-use-a-graphing-utility-to-graph-the-quadratic-function-find-the-intercepts-of-t-6

Question

In Exercises 59-64, use a graphing utility to graph the quadratic function. Find the -intercepts of the graph and compare them with the solutions of the corresponding quadratic equation when $ f(x) = 0 $.

$ f(x) = \frac{7}{10} (x^2 + 12x -45) $

Jeremy Strong · Accepted Answer

So for this question, were given the f of X is it with 7/10 times x squared plus Strom X minus 45 more ass to find the ex intercepts of the equation and then to analyze it graphically, Briefly. So what we need to do here is we need to set f of X equal to zero. Because when f of X equals zero, that means why is equal to zero. And that means that, um um, those are the locations of our ex intercepts. So we have zero equals 7/10 times X squared, strove X minus 45. Um, we can divide both sides by the 7/10 to make it go away. And we&#x27;re gonna have zero still, because he wrote about about 17th is still zero. So now we need to factor in this equation. We&#x27;re gonna have zero is Let&#x27;s see, we need to find two numbers of that multiply to give us negative 45 and they add to give us 12. So, Tim, where two numbers that set us for this. There can actually be negative 15 and positive. Three. All right, So when we&#x27;re factoring, we&#x27;re gonna need to do X um X plus 15 Tom&#x27;s ex Mona&#x27;s three. If negative 15 and three are solutions that we want the opposites of them, um, we want plus 15 and modest three within our factors to get a zero because negative 15 plus 15 0 and three modest +30 So this is our factor down equation. So we&#x27;re going to get that our ex intercepts our ex intercepts her at negative 15 zero and at 30 And if you were to graft this, you were to graft this equation. This is an upward sloping, probably because there&#x27;s no negative, no negative sign out in front. And we know that our zeroes are at three. Call this 0.3 and way out here in negative 15. So a reasonable approximation what this graph should look like. It&#x27;s something like this, something like this. And negative 15 0 and 30 hour. Our final solutions

J Hardin · Answer

let&#x27;s use a graphing utility to graft this problem. And then we&#x27;LL find the intercepts of the draft. The Ex intercepts. So here&#x27;s the graph in Dismas, and we see that we have intercepted and zero and also a four, so we can just say X equals zero X equals. For now, this is from the graphing calculator. And then now let&#x27;s find these algebraic Lee by solving the equation F of X equals zero. And the reason we&#x27;re setting into zero is because at the Ex intercepts notice that the white corner zero so setting x f events zero gives this X Square minus, for example, zero. This can be solved in many ways. Thie Easiest way is probably just a factor out the X. So we remote supplying two numbers and getting zero. This means that either X equals zero or X minus four equals zero, and solving the second equation gives us X equals zero or X equals four. So this is the algebraic, the unification of the ex intercepts that we found in the graphing calculator

J Hardin · Answer

Let&#x27;s use a graphing utility to graft this quadratic f. So here we are quadratic function. Let&#x27;s zoom out a little bit and if we zoom in, we could see what the X intercepts are. Negative two and a half and six since record those values, this is from the computer, the calculator. And now we can verify this algebraic Lee by setting the Y value to zero. So said this equal to zero and then saw for X There many ways to do this one way is to just take the left hand side and to try to factor it. So here we want numbers that multiplied to get thirty, and the one that works here is you can do plus five and then a minus six. And this last equation implies that either two X Plus five zero or X minus six zero. So solve these last two equations for X, and you get the ex intercepts. And as you can see, the solution&#x27;s agree with what we found from the consulate

J Hardin · Answer

So here. Let&#x27;s go out and graft this function F and a different window. So here&#x27;s F and here&#x27;s the graph. And just by looking at the graph, it looks like the ex intercepts or zero and five. Now let&#x27;s verify this algebraic lee. So X intercepts you get by solving this equation here so well said this equal to zero many ways to solve this one way usages. Fact throughout the eggs was maybe pull out a two x here, and we&#x27;re dealing with real number. So if you multiply and get zero, this means two X is equal to zero or negative X plus five zero equivalently X equals zero or X equals five. And this verifies that ex intercepts are the points zero zero and five zero. And you could check this another way. If you plug in zero in for X for five and Franks, they&#x27;LL make f equals zero. So this is that algebraic confirmation of the ex intercepts

J Hardin · Answer

Let&#x27;s use a graphing utility to graph the problem, ref, and to find the ex intercepts. So here&#x27;s a graphing calculator. Here is the problem, Eph, and we see that the intercepts are three and six, So you can either write the answers as X equals three and six or since intercepts or actually points, you could be more precise and write three zero comma. Six year old. Now this is from the calculator. And then let&#x27;s verify this algebraic agree by setting after zero. So setting after zero means that you&#x27;re fixing the Y value to be zero. So here&#x27;s our ethics. And then we sent that equal to zero and then solve this for X, and hopefully we obtain three or six. So there are many ways to solve this. If you notice, you can actually factor. This quadratic is X minus three and X minus six. And since we multiply, these two get zero. This implies that either X minus three zero or X minus six zero and then solving these equations for X, you get X equals three. All right, people. Six. So this is the verification of what we obtained from the graphing calculator

J Hardin · Answer

Let&#x27;s go ahead and graph F in our graphing calculator. So here&#x27;s F and it&#x27;s graft over here and read. And also we can see the ex intercepts on the graph negative to in ten. So that should say, ex intercepts. X equals minus two and ten. Or, if you&#x27;d like to be more precise, you Khun, say negative two zero or ten zero. Now this is from the graph and let&#x27;s go out and verify this algebraic lee by setting our F equals zero. So here&#x27;s F and said that equal to zero and then solve for X. So there are many ways to solve this. One way is to notice that you can factor the left hand side so this certain that shouldn&#x27;t be a six. Here, we can do X minus ten and then X plus two. So if you go ahead and foil this out, we&#x27;LL do that. Together you get X square plus two X minus ten x minus twenty. And if you simplify these middle terms, you get negative. Eight x So this equation here means one of two things here that X minus ten zero or X plus two zero equivalently. You can solve each of these last two equations for X to get X equals ten or X equals minus two just And so this is the algebraic verification over here on the right side of the ex intercepts that we obtained from the graphing calculator.

Problem

In Exercises 65-70, find two quadratic functions,…

Problem 64 Hard Difficulty

In Exercises 59-64, use a graphing utility to graph the quadratic function. Find the -intercepts of the graph and compare them with the solutions of the corresponding quadratic equation when $ f(x) = 0 $.

$ f(x) = \frac{7}{10} (x^2 + 12x -45) $

Answer

$(3,0),(-15,0)$

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$(3,0),(-15,0)$

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