Question 949638: 1.Sketch the curves of the following functions
(a) f(t)= 4t^2- 12t (b) f(x)= 4-2x-x^2 (c) f(r)=r^2-pie
2. Determine the exact least value of the expression 3x^2- 5x+2 for all real values of x/ State the exact value of x for which this least value occures. State also the range of the function f(x)= 3x^2-5x+2.
3. The graph of a quadratic function has equation of the form y=a(x-p)^2+q. Its turning point is located at (1,4). the graph also contains point (1,3).
(i) Determine the values of a, p, q.
(ii) Find the equation of the graph in the form y=ax^2+ bx+c
Please explain and teach me how i can solve these questions, it's very depressing to be the only one in the whole class who doesn't understand the whole chapter.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! we are dealing with parabolas
1) red curve is a, green curve is b, and blue curve is c
2) we are given the function f(x) = 3x^2 -5x +2
the graph of this function looks like
since the x^2 term is positive, this parabola opens upward, therefore the x value for the axis of symmetry will give us the least value
axis of symmetry is given by
x = -b / (2*a) = -(-5) / (2*3) = 5/6
now substitute for x in our function
f(5/6) = 3*(5/6)^2 -5*(5/6) +2 = -1/12
therefore the value for x is 5/6 and the least value of f(x) is -1/12
range of this function, f(x) = 3x^2 -5x +2, is f(x) > or = -1/12
3) We are given
y=a(x-p)^2+q. Its turning point is located at (1,4). the graph also contains point (1,3). This is a parabola that curves upward and the turning point tells us
i) x = -b/(2a) = 1 and
4 = a(1-p)^2 +q
the additional point tells us that
3 = a(1-p)^2 +q
The definition of a function is
"A function is a special relationship where each input has a single output", therefore this can not be a function since (1,4) and (1,3) are both on the graph of the function. Also (1,4) can not be the turning point since (1,3) is on the graph and f(x) = 3 is < f(x) = 4
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