Answer:
3. The vertex form of the function, f(x) = x² - 4·x - 17 is f(x) = (x - 2)² - 21
4. The solutions are, x = -2 + √10 and x = -2 - √10
5. The quadratic equation with vertex (3, 1) and a = 1 in standard form is given as follows;
f(x) = x² - 6·x + 10
Explanation:
3. The function given in standard form is f(x) = x² - 4·x - 17, which is the form, f(x) = a·x² + b·x + c
The vertex form of the of a quadratic function can be presented based on the above standard form as follows;
f(x) = a(x - h)² + k
Where;
(h, k) = The coordinate of the vertex
h = -b/2a
k = f(h)
Comparing with the given equation, we have;
f(x) = a·x² + b·x + c = x² - 4·x - 17
a = 1
b = -4
c = -17
∴ h = -(-4)/(2 × 1) = 2
h = 2
k = f(h) = f(2) = 2² - 4 × 2 - 17 = -21
k = -21
The vertex form of the function, f(x) = x² - 4·x - 17 is therefore, given as follows;
f(x) = (x - 2)² - 21
4. The given equation for which we need to solve by completing the square is 2·x² + 8·x = 12
Dividing the given equation by 2 gives;
x² + 4·x = 6
Which is of the form, x² + b·x = c
Where;
a = 1
b = 4
c = 6
From which we add (b/2)² to both sides to get x² + b·x + (b/2)² = c + (b/2)²
Adding (b/2)² = (4/2)² to both sides of x² + 4·x = 6 gives;
x² + 4·x + 4 = 6 + 4
(x + 2)² = 10
x + 2 = ±√10
x = -2 ± √10
The solution are, x = -2 + √10 and x = -2 - √10
5. Given that the value of the vertex = (3, 1), and a = 1, we have;
The vertex, (h, k) = (3, 1)
h = 3, k = 1
Therefore, h = 3 = -b/(2 × a) = -b/(2 × 1)
∴ -b = 2 × 3 = 6
b = -6
k = f(h) = a·h² + b·h + c, by substitution, we have;
k = f(3) = 1 × 3² + (-6) × 3 + c = 1
∴ c = 1 - (1 × 3² + (-6) × 3) = 10
c = 10
The quadratic equation with vertex (3, 1) and a = 1 in standard form, f(x) a·x² + b·x + c is therefor;
f(x) = x² - 6·x + 10