Help with these algebra problmes please?

Question

find intercepts of f(x)= 6x + x(x-1) the x intercepts are=

2. give exact and approximate solutions to three decimal places x^2- 5x +3=0 the solutions for x are=

the approximate solution to 3 decimals places are x=

3. f(x)=-x^2 + 10x +3 the vertix = line symmerty= the min/max of f(x)is=

4. solve for x x^2 + 41= 8x the solutions are=

Answer 1

1) Solve f(x) = 0 for x to find the x-intercept:

f(x) = 0

6x + x(x - 1) = 0

We factorise out the x:

x(6 + (x - 1)) = 0

x(x + 5) = 0

x = 0 or -5

We have two x-intercepts: (0,0) and (-5,0).

2) This one is best solved using the quadratic formula:

x^2 - 5x + 3 = 0

x = [-b +/- sqrt(b^2 - 4ac)] / [2a]

x = [5 +/- sqrt(5^2 - 4 * 1 * 3] / [2 * 1]

x = [5+/- sqrt(13)] / 2

x = (5 + sqrt(13)) / 2 or (5 - sqrt(13)) / 2

x = 4.303 or 0.697 (3 dp)

3) The vertex of a parabola lies on the axis of symmetry, and the minimum/maximum value are the function value at the vertex. The axis of symmetry is given by this formula:

x = -b / [2a]

You may notice that we can get this from splitting up the quadratic formula:

x = [-b +/- sqrt(b^2 - 4ac)] / [2a]

x = -b / [2a] +/- sqrt(b^2 - 4ac) / [2a]

Notice that in the middle of the two values you get it x = -b / [2a], which makes sense, because it's the axis of symmetry!

Anyway, in this parabola, the axis of symmetry is:

x = -10 / [2*-1]

x = 5

The quadratic has a negative x^2 coefficient, so the parabola will be upside-down, and therefore have a maximum value. That maximum value will be the function value at x = 5.

f(5) = -(5)^2 + 10(5) + 3

f(5) = 28

So 28 is the maximum value. Also, the vertex will be the point (5,28).

4) x^2 + 41 = 8x

x^2 - 8x + 41 = 0

You can solve this using the formula, but I'm going to complete the square:

x^2 - 8x + (8/2)^2 - (8/2)^2 + 41 = 0

x^2 - 8x + 16 - 16 + 41 = 0

x^2 - 8x + 16 + 25 = 0

(x - 4)^2 + 25 = 0

(x - 4)^2 = -25

Now we stop, because we can't find the square root of a negative number. There are no real solutions.

Answer 2

1. f(x)= 6x + x^2 -x

f(x) = 5x + x^2

f(x) = x(5 +x)

x = 0

5 +x = 0

x= -5

2. (5+sqrt(25+12))/2 =5.541

(5-sqrt(25+12))/2= -0.541

3. I had to use differentiation but the answer is x=5, f(x)=(28)

4.(8+sqrt(64-164))/2 =

(8-sqrt(64-164))/2 =

No real roots. They are imaginary since we cannot get the square root of a negative number.

Answer 3

1) To find the X intercept, you must set f(x) to 0.

0 = 6x + x(x-1)

Distribute:

0 = 6x + x^2 - x

Simplify:

0 = x^2 +5x

Simplify:

0 = x(x+5)

So either x = 0 or x + 5 = 0

The solution is (0,-5)

Make sense?

2) Use the Quadratic Equation with A = 1, B = -5 and C = 3

3) X Value of Vertex = -B/2A = -10/ 2 = -5

Y value of vertex = (-5)^2 + 10(-5) + 3 = 25-50+3 = -22

The Vertex = (-5,-22)

The Line of symmetry is x = -5

The min is -22

There is no max.

4)  Use quadratic equation where A=1, B=-8 and C=41