# Find an equation of the set of all points equidistant from the points [Solved]

Question 1

Question 2

Question 3

The equation of a parabola is y = x 2 − 4 x − 5. Express this equation in factored form and state the x-intercepts. Parts of a Parabola: To define the characteristics of a parabola, an equation would be the best way to express it. However, it is also defined by its vertex, focus and axis of symmetry.
Question 4

Let f ( x ) = 5 − 4 x 3 . (a) Find a simplified form of the difference quotient f ( x + h ) − f ( x ) h . (b) Complete the following table x h f ( x + h ) − f ( x ) h 3 2 3 1 3 0.1 3 0.01 Difference Quotient: The difference quotient of a function, f ( x ) gives the slope of a secant line which is passing through two points on the curve. This is calculated using: f ( x + h ) − f ( x ) h Note: By applying the limit as h → 0 , we get the derivative of the function.

The parabola’s equation is y=x2−4x−5y=x2−4x−5. It has to be factored in order to get the x-intercepts.

(x−5)(x+1)=0(x−5)(x+1)=0

To determine the x-intercepts, the equation has to be equated to zero. Using the factors, equating them to zero and determining the value of xx:

x−5=0x=5andx+1=0x=−1x−5=0x=5andx+1=0x=−1

Thus, the x-intercepts are (5,0) and (-1, 0).

The given function is:

f(x)=5−4x3f(x)=5−4×3

(a) Replace x by x+h in the above function:

f(x+h)=5−4(x+h)3=5−4×3−12x2h−12xh2−4h3f(x+h)=5−4(x+h)3=5−4×3−12x2h−12xh2−4h3

The difference quotient is:

f(x+h)−f(x)h=(5−4×3−12x2h−12xh2−4h3)−(5−4×3)h=5−4×3−12x2h−12xh2−4h3−5+4x3h=−12x2h−12xh2−4h3h=h(−12×2−12xh−4h2)h=−12×2−12xh−4h2f(x+h)−f(x)h=(5−4×3−12x2h−12xh2−4h3)−(5−4×3)h=5−4×3−12x2h−12xh2−4h3−5+4x3h=−12x2h−12xh2−4h3h=h(−12×2−12xh−4h2)h=−12×2−12xh−4h2

(b) (i) When x=3;h=2x=3;h=2, the difference quotient is:

−12(3)2−12(3)(2)−4(2)2=−196−12(3)2−12(3)(2)−4(2)2=−196

(ii) When x=3;h=1x=3;h=1, the difference quotient is:

−12(3)2−12(3)(1)−4(1)2=−148−12(3)2−12(3)(1)−4(1)2=−148

(iii) When x=3;h=0.1x=3;h=0.1, the difference quotient is:

−12(3)2−12(3)(0.1)−4(0.1)2=−111.64−12(3)2−12(3)(0.1)−4(0.1)2=−111.64

(iv) When x=3;h=0.01x=3;h=0.01, the difference quotient is:

−12(3)2−12(3)(0.01)−4(0.01)2=−108.3604

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