Find the minimum or maximum value of each function. then State the domain and range of the function. 1. g(x)=x^2-2x+1 2.h(x)= 5x^2+ 15x-3

1.

The minimum value of g(x) is (1 , 0)

The domain of g(x) is {x : x ∈ R}

The range of g(x) is {y : y ≥ 0}

2.

The minimum value of h(x) is (-1.5 , -14.25)

The domain of h(x) is {x : x ∈ R}

The range of h(x) is {y : y ≥ -14.25}

Step-by-step explanation:

The quadratic function f(x) = ax² + bx + c is represented graphically by a parabola which has a minimum/maximum vertex (h , k), where

Now let us solve the problems

1.

∵ g(x) = x² - 2x + 1

∵ The form of the quadratic function is f(x) = ax² + bx + c

∴ a = 1 , b = -2 , c = 1

∵ a is positive

∴ The function has a minimum value

∵ h =

∴ h =

- To find k substitute x in g(x) by 1

∴ g(1) = (1)² - 2(1) + 1 = 1 - 2 + 1

∴ g(1) = 0

∵ k = g(h)

∴ k = 0

∴ The minimum value of g(x) is (1 , 0)

∵ The domain of the quadratic function is all real numbers

∴ The domain of g(x) is {x : x ∈ R}

∵ The range of the quadratic function with minimum value is

   {y : y ≥ k}

∴ The range of g(x) is {y : y ≥ 0}

2.

∵ h(x) = 5x² + 15x - 3

∴ a = 5 , b = 15 , c = -3

∵ a is positive

∴ The function has a minimum value

∵ h =

∴ h =

∴ h = - 1.5

- To find k substitute x in h(x) by 1

∴ h(-1.5) = 5(-1.5)² + 15 (-1.5) - 3 = 11.25 - 22.5 - 3

∴ h(-1.5) = -14.25

∵ k = h(h)

∴ k = -14.25

∴ The minimum value of h(x) is (-1.5 , -14.25)

∵ The domain of the quadratic function is all real numbers

∴ The domain of h(x) is {x : x ∈ R}

∵ The range of the quadratic function with minimum value is

   {y : y ≥ k}

∴ The range of h(x) is {y : y ≥ -14.25}

i can you with this question sweetie!

step-by-step explanation:

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3