Final answer:
a) To show that f(x)>0 for all real values of x, we need to prove that the quadratic function f(x)=x^2-8x+18 has no real roots. b) Express f(x) in the form f(x)=(x+p)^2+q. c) The equation of the parabola's axis of symmetry is x=-p, and the coordinates of the vertex are (4,1). d) The range of values of k for the equation 2x^2+(3-k)x+k+3=0 to have two distinct real solutions is k<-3 or k>13.
Step-by-step explanation:
a) Proving f(x) > 0:
To prove that f(x) > 0 for all real values of x, we need to show that the quadratic function f(x) = x^2 - 8x + 18 has no real roots. We can determine this by finding the discriminant.
The discriminant of a quadratic function ax^2 + bx + c is given by the formula D = b^2 - 4ac. If the discriminant is negative, the quadratic function has no real roots.
For f(x) = x^2 - 8x + 18, the discriminant is D = (-8)^2 - 4(1)(18) = 64 - 72 = -8. Since the discriminant is negative, the quadratic function has no real roots, which means that f(x) > 0 for all real values of x.
b) Expressing f(x) in the form f(x) = (x + p)^2 + q:
To express f(x) in the form f(x) = (x + p)^2 + q, we need to complete the square. The general form of completing the square for a quadratic function ax^2 + bx + c is (x + (b/2a))^2 - (b^2 - 4ac)/4a.
For f(x) = x^2 - 8x + 18, we have a = 1, b = -8, and c = 18. Plugging these values into the completing the square formula, we get f(x) = (x + (-8/2))^2 - ((-8)^2 - 4(1)(18))/(4*1) = (x - 4)^2 - (64 - 72)/4 = (x - 4)^2 + 1.
c) Equation of the parabola's axis of symmetry and coordinates of its vertex:
The equation of the parabola's axis of symmetry is x = -p, where p is the x-coordinate of the vertex. In this case, since the vertex is (4, 1), the equation of the parabola's axis of symmetry is x = -4.
The coordinates of the vertex can be found by setting x = -b/2a in the quadratic function. For f(x) = x^2 - 8x + 18, the x-coordinate of the vertex is -(-8)/(2*1) = 8/2 = 4. Plugging this x-coordinate into the quadratic function, we get f(4) = (4 - 4)^2 + 1 = 0 + 1 = 1. So, the coordinates of the vertex are (4, 1).
d) Range of values of k for two distinct real solutions:
The quadratic equation 2x^2 + (3 - k)x + k + 3 = 0 has two distinct real solutions if the discriminant D is greater than 0. The discriminant is given by D = (3 - k)^2 - 4(2)(k + 3).
To find the range of values of k, we can set the discriminant greater than 0 and solve for k. (3 - k)^2 - 4(2)(k + 3) > 0. Expanding and simplifying, we get k^2 - 10k - 39 > 0.
Factoring the quadratic inequality, we get (k - 13)(k + 3) > 0. The values of k that satisfy this inequality are k < -3 or k > 13.