# Solving Problems With Quadratic Functions The solution to an equation is sometimes referred to as the root of the equation.An important theorem, which cannot be proved at the level of this text, states "Every polynomial equation of degree n has exactly n roots." Using this fact tells us that quadratic equations will always have two solutions. The simplest method of solving quadratics is by factoring.Let x be the number of price increases of 10 cents Let y be the revenue a) Revenue = Price x Quantity y = (2.00 0.10x)(300 – 25x) b) The zeros will be when 2.00 0.10x = 0 and 300 – 25x = 0 x = -20 and x = 12 The midpoint is x = -4 There the optimal price will be 4 price decreases.

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For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship.

Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity.*Identify the effect on the graph of replacing f(x) by f(x) k, k f(x), f(kx), and f(x k) for specific values of k (both positive and negative); find the value of k given the graphs.

The time is less than double the previous time due to acceleration. so in factored form the equation will be y = a(x – 0)(x – 10) Substitute in known point (2, 20) 20 = a(2 – 0)(2 – 10) 20 = -16a a = -1.25 [note it makes sense that a is negative, since parabolas opens down] Equation is y = - 1.25(x)(x – 10) The great height will be at 5 s which is the midpoint of the two zeros 0 and 10.

b) To find when he hits the water set H = 0 and solve 0 =-4.9t2 21 0 =-4.9t 2 21 4.9t 2 = 21 21 4.9 t  2.07 t  Time must be positive, so the diver hits the water in about 2.1 seconds.

So if this is the height, the ground is when the height is equal to 0. So if we apply it, we get t is equal to negative b. So negative negative 10 is going to be positive 10. And this over here, we have a-- let's see if we can simplify this a little bit. And this is equal to 10 plus or minus-- square root of 900 is 30-- over 16. Or t is equal to 10 minus 30, which is negative 20 over 16.

So hitting the ground means-- this literally means that h is equal to 0. Plus or minus the square root of negative 10 squared. The negative sign, negative times a negative, these are going to be positive. And so we get time is equal to 10 plus 30 over 16, is 40 over 16, which is the same thing if we divide the numerator and denominator by 4 to simplify it as 10 over-- or actually even better, divide it by 8-- that's 5 over 2. Divide the numerator and the denominator by 4, you get negative 5 over 4. A border of flowers are planted around the garden so that the area of the flowers is exactly one half the area of the field. The equation area of the garden will be Area = (30 – 2x)(40 – 2x) Area needs to be ½ the area of the field so will be 600m2 600 = (30 – 2x)(40 – 2x) 600 = 1200 – 140x – 4x2 4x2 140x – 600 = 0 (divide by 4) x2 35x – 150 = 0 (factor) (x 30)(x 5) = 0 so x = -30 or x= 5 since distance is positive x = 5. Give students problems that allow for multiple points of entry and let them use lots of time and technology to develop their own methods to solve them--after several days, you will be excited to see all the different methods they come up with!You now have the necessary skills to solve equations of the second degree, which are known as quadratic equations.Upon completing this section you should be able to: bx c = 0 when a ≠ 0 and a, b, and c are real numbers.All quadratic equations can be put in standard form, and any equation that can be put in standard form is a quadratic equation.In other words, the standard form represents all quadratic equations.b) Find the optimum price for a salad c) What is the optimal revenue?Solution: Assume that when the price decreases by 10 cents they will sell 25 more salads.Experiment with cases and illustrate an explanation of the effects on the graph using technology.Include recognizing even and odd functions from their graphs and algebraic expressions for them.

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