How will you draw a flow chart to find the roots of a quadratic equation?
input a,b,c (the coeffs in a*x^2+b*x+c=0)if a=0 then return “error, not a quadratic equation”compute temp=b^2–4*a*cif temp>=0 then compute temp2=sqrt(temp) else compute temp2=sqrt(-temp)if temp<0 then return “Complex conjugate roots are” (-b+temp2*%i)/(2*a) “and”(-b-temp*%i)/(2*a) elseif temp=0 then return “double root” -b/(2*a) else return “roots are”(-b+temp2)/(2*a) “and” (-b-temp2)/(2*a)endClose enough for a flowchart in words?
Math Question, Quadratic Formula?
That equation is: h = - 4.9t² + 8.4t + 1.5, right? You have left out t in 8.4t. You can do this two ways, one is pure numerical method. Chart it and you will find the maximum height. Otherwise, you can differentiate it and set it to zero. Since at maximum or minimum value, the slope is zero, you can find the value of t at which h is maximum. Differentiating, -9.8t +8.4 = 0, or t = 8.4/9.8 = 0.857 second. Plug this value of t in the equation for h and you get maximum height.
I just don't get Quadratic Functions/Graphs! IM SO CONFUSED! please help?
Well, first it says to graph y = -1/2x^2 – 1. I'm not really sure how to do that. I'm homeschool, and I JUST started this unit, and it's already telling me to graph this, but it hasn't even explained it. And then it says to use that graph to identify the vertex, and tell whether its a minimum or maximum. Please, I'm not trying to get answers, I want HELP! Step by step help for begining with quadratic functions/graphs. Please, I'm getting really discouraged.
How can I transfer a trendline equation from the graph to the cells where I can use it for further calculations?
Hi!I only know how to automate linear trendlines so this will be the answer scope here.A brief intro:The linear trendline equation is made of two excel functions, linked to your dataset.Let me break it in steps:1. The form of this equation follows the linear standard: y = ax + b2. You may have noticed that the y and the x are already displayed in your chart, so I will explain the a and b from the equation above as to how they connect to the functions I have mentioned in the intro: "a" is your slope, the inclination of your trend lineThis can be automated by using the formula slope(y,x) , where y represents your dependend variable data column (yes, all vertical axis inputs in your chart) and x is all your horizontal inputs.beware: if you are using sales vs any time series, do not use the value of the time series like "2014","2015", etc - use 1, 2, 3, etc as if you were counting your y inputsWhy? Excel does it if you do not define any data column as your x parameter."b" is your intercept, defining how high in y is your trend line, with the same parameters as above.3. Plot the formula intercept(y,x) + slope(y,x) * your input This will use the same trendline displayed in your chart, simulating new values, or checking old ones.Hope it helps
PreCal Quadratic Function?
Let x = # rooms rented. C(x) = cost function, which is the cost to service each room per day. C(x) = 10x The price to rent each room is a linear function. If you make a chart, it might help to find the equation of this line. Price # rooms 80 300 81 297 82 294 From this, the slope of the line, m, is -3. Using the y=mx+b, plug in what is known and find the b. 300 = -3(80) + b (The 300 and 80 come from the chart above and are plugged into y=mx+b. Solving for b, b=540. Our price function, p is p = -3x+540, where x is still the number of rooms rented. This tells us the price to rent the rooms for in order to rent x number of rooms. The revenue function, R(x), is the price per room times the number of rooms rented. R(x) = x*p = x(-3x+540) R(x) = -3x^2 + 540x The profit, P(x), is the difference between the revenue and the cost. P(x) = R(x) - C(x) P(x) = -3x^2 + 540x - 10x P(x) = -3x^2 + 530x The graph of this is a parabola that opens downward. The maximum profit occurs at the vertex. Use -b/2a to find the x value of the maximum profit. x = -530/(2*-3) x = 88.33 rooms, or just 88 rooms. Now find P(88). P(88) = -3*88^2 + 530*88 P(88) = $23,408 The price per room is p = -3x + 540 For 88 rooms: p = -3*88 + 540 p = $276
How do I figure this out, in equation form?
Ok. Let x = the original number of students. The whole cost was $160. Since there are several students, they split the cost evenly among them. The price each student paid after splitting the cost is represented by: 160/x. Now, since x is the original number of students, when one joins them, the new expression will be x+1. Just as we did before, the price each student paid after splitting the cost is represented by: 160/(x+1). Now, here comes the weird part. We said that if another student joints the group, each now pays $8.00 LESS than if the student didn't join. Let's recap what we have. Original cost for each student: 160/x. New cost for each student: 160/(x+1). We know that the new price is $8.00 less so we can set up an equation: 160/x - 8 = 160/(x+1). Now let's solve for x. We multply 8 by (x/x) so we can get a common denominator as follows: 160/x - 8(x/x) = 160/(x+1) 160/x - 8x/x = 160/(x+1) We can now put 160 and 8x together (160 - 8x)/x = 160/(x+1) We now cross multiply to get: 160x + 160 -8x^2 -8x = 160x Subtract 160x from both sides: 160- 8x^2 -8x = 0 Arrange in decreasing power form: -8x^2 -8x + 160 = 0 Now, multiply by -1 to get the x^2 term positive: 8x^2 + 8x - 160 = 0 Factor out an 8: 8 (x^2 + x - 20) = 0 Factor the quadratic: 8(x - 4) (x + 5) = 0 Make a t-chart: 8 is not equal to zero. For x - 4, we get x - 4 = 0 Add 4. x = 4. For x + 5, we get x + 5 = 0 Subtract 5 x = -5 Now, we get x =4 and x = -5. Since you can't have a negative number of students, we reject -5, so the answer is 4 students in the group originally!
How to graph a parabola?
the best way is to start by finding the vertex. the vertex in standard form (y=ax^2+bx+c) can be found by first finding the x value of this vertex (-b/2a) then sub that in 4 the y value. after ploting the vertex, plug in values for x and find the corresponing y values. Plot until u get the idea of the graph.
What is the difference between linear and quadratic equations?
Linear equations are generally in this format:2x - 4 = yAnd when graphing linear equations: (PS the equation and the graph below do not match - just a general format of linear equations’ graph)Whereas quadratic equations are in this type of format:x^2 + 5x + 6 = yAnd when graphing quadratic equations: (again, the equation and graph do not match)***So basically, when we say linear equations, we mean those equations that have a maximum of only one x-intercept, or in other words, when y = 0, they only have a maximum of one coordinate that passes through the x-axis -> as seen in the graph above.However, quadratic equations are a little bit different. Whereas linear equations have only a maximum of one x-intercept, quadratic equations’ graph (a parabola) has a maximum of two x-intercepts, meaning that when y = 0, the parabola passes through a maximum of two coordinates in the x-axis -> as seen in the graph above.PS- A parabola and a linear equation graph don’t have any minimum of points: for ex: a linear equation graph can have 0 x-intercepts, and a parabola can have 1 or even no x-intercept at all (Just some general information that I thought you might want to know for greater understanding).