Guide :

# what is the quadratic standard form of f(x)=2(x-6)(20-3x) ?

what are the steps?

## Research, Knowledge and Information :

### Quadratic Functions in Standard Form - analyzemath.com

Quadratic Functions in Standard Form. Quadratic functions in standard form. f(x) = a(x - h) 2 + k and the properties of their graphs such as vertex and x and y ...

### Quadratic Equations - Math is Fun

An example of a Quadratic Equation: Quadratic Equations make nice ... In Standard Form a, b and c; x 2 = 3x − 1: ... Quadratic Equation in Standard Form: ax 2 + bx ...

### Quadratic Function in Standard Form

Quadratic Function in Standard Form. ... Given quadratic function is f(x) = 3x 2 + 6x + 12. Comparing with the general form of the quadratic function, f(x) ...

A quadratic function is one of the form f(x) = ax 2 + bx + c, where a, b, and c are numbers with a not equal to zero. ... Write f(x) = 3x 2 + 12x + 8 in standard form.

### Quadratic Equation Solver - Math Is Fun

Quadratic Equation Solver. ... Is it Quadratic? Only if it can be put in the form ax 2 + bx + c = 0, ... In standard form a, b and c; x 2 = 3x -1:

### Chapter 3.2 Quadratic Functions - JoeMath.Com

Write a Quadratic Function in Standard Form f(x) = x 2 + 6x + 8: ... = x 2 + 6x + 9-9 + 8 Take half of 6 and square it: f(x) = (x + 3) 2-1 ... 2 + 20 (x - 5) 2 = x 2 ...

### MATH 11011 EXPRESSING A QUADRATIC FUNCTION KSU IN STANDARD FORM

Quadratic function in standard form, page 3 4. f(x) = 2x2 +8x¡1 Before we can complete the square, we need to factor a 2 from each x term. f(x) = 2x2 +8x¡1

Rewrite the equation 6x^2 + 3 = 2x - 6 in standard form and identify a, b, and c. So standard form for a quadratic equation is ax squared plus bx plus c is equal to ...

### Quadratic Functions(General Form) - analyzemath.com

Examples of quadratic functions. f(x) = -2x 2 + x - 1; f(x) = x 2 + 3x + 2; ... This is the standard form of a quadratic function with h = ...

### Quadratic: Standard Form - Edmonton Public Schools

Quadratic Standard (Completed Square) Form Interactive Activity. ... Slider b will not be studied in Math 20 Pure. ... the equation of the quadratic is y = 2(x - 2) ...

## Suggested Questions And Answer :

### Can you simplify it step by step as well

Answered earlier: 26r-2=3r^2. We need to put this into standard form. The standard quadratic form is ax^2+bx+c=0, where x is the unknown and a, b and c are numbers. The unknown is r in this problem, rather than x, so let's put the equation into quadratic form: 3r^2-26r+2=0. How did I get this? Subtract 26r from each side: -2=3r^2-26r. Now, add 2 to each side: 0=3r^2-26r+2, which is the same as 3r^2-26r+2=0. (There's no mystery here. The equals sign shows us that we have equality of quantity. If, for example, an apple cost 30 cents or pennies, then 30 cents will buy an apple, so we can say apple=30 or 30=apple.) This equation doesn't factorise, so we need the formula to solve it: r=(26+sqrt(26^2-4*3*2))/6 (this is the quadratic formula x=(-b+sqrt(b^2-4ac))/2a. All I've done is put b=-26, a=3 and c=2 into the formula. The plus-or-minus sign (+) means there are two answers, one using plus and the other using minus.) So r=(26+sqrt(676-24))/6=(26+sqrt(652))/6=8.589 or 0.0776. Because b=-26, -b=26: that's why the minus has disappeared. I hope the extra info helps you to understand.

### quadratic function written in standard form solutions (5,2) (0,2) (8,-6)

y=ax^2+bx+c. When x=0, y=c=2. (5,2): 2=25a+5b+2; (8,-6): -6=64a+8b+2. 25a+5b=0, so b=-5a; 64a+8b=-8; 8a+b=-1; substituting for b: 8a-5a=-1, so a=-1/3 and b=5/3. The equation can be written y=-x^2/3+5x/3+2 (standard form) or 3y=6+5x-x^2.

### how can I simplify the last problem

26r-2=3r^2. We need to put this into standard form. The standard quadratic form is ax^2+bx+c=0, where x is the unknown and a, b and c are numbers. The unknown is r in this problem, rather than x, so let's put the equation into quadratic form: 3r^2-26r+2=0. How did I get this? Subtract 26r from each side: -2=3r^2-26r. Now, add 2 to each side: 0=3r^2-26r+2, which is the same as 3r^2-26r+2=0. (There's no mystery here. The equals sign shows us that we have equality of quantity. If, for example, an apple cost 30 cents or pennies, then 30 cents will buy an apple, so we can say apple=30 or 30=apple.) This equation doesn't factorise, so we need the formula to solve it: r=(26+sqrt(26^2-4*3*2))/6 (this is the quadratic formula x=(-b+sqrt(b^2-4ac))/2a. All I've done is put b=-26, a=3 and c=2 into the formula. The plus-or-minus sign (+) means there are two answers, one using plus and the other using minus.) So r=(26+sqrt(676-24))/6=(26+sqrt(652))/6=8.589 or 0.0776. Because b=-26, -b=26: that's why the minus has disappeared. I hope the extra info helps you to understand.

26r-2=3r^2. The standard quadratic form is ax^2+bx+c=0, where x is the unknown and a, b and c are numbers. The unknown is r in this problem, so let's put the equation into quadratic form: 3r^2-26r+2=0. Note the sign changes as terms are transferred from one side to another, and also note that, in order to preserve the 3r^2 on the right I've brought all the terms over to the right to bring them all together. In doing so, which would have made the equation 0=3r^2-26r+2, I've moved the equals and zero over to the right to get it into quadratic form. I can do this because if A=B then it's true that B=A. This equation doesn't factorise, so we need the formula to solve it: r=(26+sqrt(26^2-4*3*2))/6 (this is the quadratic formula x=(-b+sqrt(b^2-4ac))/2a). So r=(26+sqrt(676-24))/6=(26+sqrt(652))/6=8.589 or 0.0776.

### write this Quadratic Equation in Standard Form : (2x-1)^2=(x+1)^2

(2x-1)^2=(x+1)^2 4x^2-4x+1=x^2+2x+1 3x^2-6x=0 is standard form (but constant is zero). This is simplified by dividing through by 3: x^2-2x=0 Another way: (2x-1)^2-(x+1)^2=(2x-1+x+1)(2x-1-x-1)=3x(x-2)=3x^2-6x=x^2-2x=0 (the solutions are x=0 and x=2).

A quadratic equation in the standard form is given by ax 2 + bx + c = 0 where a, b and c are constants with a not equal to zero. ...Solve the above equation to find the quadratic fomulas Given ax 2 + bx + c = 0 Divide all terms by a x 2 + (b / a) x + c / a = 0 Subtract c / a from both sides x 2 + (b / a) x + c / a - c / a = - c / a and simplify x 2 + (b / a) x = - c / a Add (b / 2a) 2 to both sides x 2 + (b / a) x + (b / 2a) 2 = - c / a + (b / 2a) 2 to complete the square [ x + (b / 2a) ] 2 = - c / a + (b / 2a) 2 Group the two terms on the right side of the equation [ x + (b / 2a) ] 2 = [ b 2 - 4a c ] / ( 4a2 ) Solve by taking the square root x + (b / 2a) = ± sqrt { [ b 2 - 4a c ] / ( 4a2 ) } Solve for x to obtain two solutions x = - b / 2a ± sqrt { [ b 2 - 4a c ] / ( 4a2 ) } The term sqrt { [ b 2 - 4a c ] / ( 4a2 ) } may be written sqrt { [ b 2 - 4a c ] / ( 4a2 ) } = sqrt(b 2 - 4a c) / 2 | a | Since 2 | a | = 2a when a > 0 and 2 | a | = -2a when a < 0, the two solutions to the quadratic equation may be written x = [ -b + sqrt( b 2 - 4a c ) ] / 2 a x = [ -b - sqrt( b 2 - 4a c ) ] / 2 a The term b 2 - 4a c which is under the square root in both solutions is called the discriminant of the quadratic equation. It can be used to determine the number and nature of the solutions of the quadratic equation. 3 cases are possible case 1: If b 2 - 4a c > 0 , the equation has 2 solutions. case 2: If b 2 - 4a c = 0 , the equation has one solutions of mutliplicity 2. case 3: If b 2 - 4a c < 0 , the equation has 2 imaginary solutions.

In a quadratic expression in x, c is the value of the expression when x=0; in a linear equation in x, b, the y intercept, is the value when x=0. So the quadratic and linear expressions both reduce to the constant term (b or c) when  x=0, and both are points along the y axis when the equations are plotted.

### Quadratic Functions, Dividing Polynomials, and Zeros of Polynomials

1. The maximum area is when the rectangle is a square. So the three fenced sides are equal and have length 150/3=50 yd. Let's look at the logic: the sides of the rectangle are L and W so the area, A, is LW and the perimeter, P=2L+2W. L=(P-2W)/2 and A=W(P-2W)/2=(1/2)(WP-2W^2). The maximum value of A is obtained by differentiating the quadratic with respect to W: (1/2)(P-4W)=0 at a turning point, so W=P/4 and L=(P-P/2)/2=P/4. Therefore L=W=P/4. The enclosed area is therefore a square, and we know that the three sides plus another unfenced length, L, make up the perimeter P=4L=150+L, so 3L=150 and L=50 yd. Alternative solution avoiding calculus: a. A=LW where L=length of rectangular area, W=width. b. 150=2L+W. c. W=150-2L. But the true perimeter of the area is P=2L+2W, so W=(1/2)(P-2L). d. A=L(P-2L)/2=(LP-2L^2)/2=LP/2-L^2. e. f(L)=-L^2+LP/2-A in standard quadratic form f(x)=ax^2+bx+c, where a=-1, b=P/2, c=-A and x=L. f. -b/2a=(-P/2)/(-2)=P/4. This is the vertex of f(L), and L=P/4 is the value for the maximum area. g. From (c) W=(1/2)(P-2L)=P/4, so W=L=P/4. This makes the area a square of area P^2/16. From (b) 150=2L+W=P/2+P/4=3P/4, making P=4/3*150=200, so W=L=200/4=50 yd. h. Maximum area is 50^2=2500 sq yd. 2. Synthetic division (1) -3 | 2 -3 -45 -54 ......2 -6..27...54 ......2 -9 -18 | 0 3. Synthetic division (2) 6 | 2..-9 -18 .....2 12..18 .....2...3..| 0 This quotient is 2x+3, representing the third factor of the polynomial. 4. Synthetic division (3) -4 | 2 -3 -45..-54 ......2 -8..44.....4 ......2 -11 -1 | -50 The Remainder Theorem tells us that substituting x=-4 into the polynomial gives us the same remainder. The substitution gives: -128-48+180-54=-50. The zeroes are -3, 6 and -3/2, because synthetic division (2) tells us that 2x+3 is a factor.

### y=4x^2+12x-11

x-coordinate of the vertex= -b/(2a). "b" is the second coefficient and "a" is the first. Once you find this, plug that in for x and solve for y. The vertex will be this point, (x,y). y=4x^2+12x-11. x-coordinate= -b/(2a)= -12/(2*4)= -12/8. Simplied: -3/2. Plug in x= -3/2 and solve for y. y=4(-3/2)^2+12(-3.2)-11.  ---->           y= 4(9/4)+12(-3/2)-11.  ---->         y= 9-18-11 ----->  y=-20 VERTEX: Ans. (-3/2, -20) POINT OF SYMMETRY: This is either the same thing as the vertex, or if you needed the AXIS OF SYMMETRY, that would be x= -3/2 (a vertical line going through the vertex, dividing the parabola symmetrically in half). Y-INTERCEPT: This is easy when quadratics are i standard form (not vertex or factored form). You plug in x=0 and solve for y. This is easier in standard form like how the quation was given above, because plugging in x=0 gets rid of any term except the last term. Thus, the y-coordinate is equal to the last term. The answer is to be written as a point: (0, -11). Good luck and I hope this helps you