I don't remember how to do these ones. can someone help me as much as possible? thanks!
I have a test comeing up for finals and owul like to be able to do these without hesitation...
If the coefficient of the quadratic term is negative, the parabola opens downward. That eliminates choice A as an answer.
Set f(x)=0, and solve for x to find the zeros of the function (where the parabola crosses the x-axis)
$\displaystyle -3x^2+7x-4=0$
$\displaystyle 3x^2-7x+4=0$
$\displaystyle (3x-4)(x-1)=0$
$\displaystyle x=\frac{3}{4} \ \ or \ \ x=1$
That looks like choice B.
The x-coordinate of the vertex is $\displaystyle \frac{-b}{2a}$ in the general form $\displaystyle ax^2+bx+c=0$
$\displaystyle x=\frac{-7}{-6}=\frac{7}{6}$
To find the y-coordinate of the vertex, find f(7/6) in your original function $\displaystyle f(x)=-3x^2+7x-4$
You have to take the original function $\displaystyle f(x)=x^2+10+27$ and complete the square in order to convert it to vertex form. Are you familiar with "completing the square"?
Step 1: Take half the coefficient of x and square it. That would be 1/2 of 10, which is 5 and 5 squared = 25. That's where I get the 25. I must add that value to the function in order to make $\displaystyle x^2+10+25$ a perfect square trinomial; namely $\displaystyle (x+5)^2$
Step2: I can't forget about the 27 I already had, and I must subtract the 25 that I added to make my perfect square trinomial. Now everything is balanced again.
$\displaystyle f(x)=x^2+10x+5^2 \ \ +27 \ \ - 25$
$\displaystyle f(x)=(x+5)^2+2$
This is now in the form: $\displaystyle f(x)=a(x-h)^2+k$ where (h, k) is the vertex of the parabola.