Answer:
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
- h = [tex]-\frac{b}{2a}[/tex] and k is the value of f(h)
- The minimum/maximum value of the function is (h , k)
- The domain of the function is {x : x ∈ R} (all real numbers)
- The range of the function is {y : y ≥ k} if the function has a minimum value OR {y : y ≤ k} if the function has a maximum value
- If a (coordinate x²) is positive, then the function has a minimum value if a is negative, then the function has a maximum value
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 = [tex]-\frac{b}{2a}[/tex]
∴ h = [tex]-\frac{(-2)}{2(1)}=1[/tex]
- 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 = [tex]-\frac{b}{2a}[/tex]
∴ h = [tex]-\frac{(15)}{2(5)}=-\frac{15}{10}[/tex]
∴ 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}
