Question

Find the y-intercept of the following function. Enter your answer as a number only, like this example: -12.
g(x) = - x2 + 6x – 8

Answer 1

to find y-intercept let x=0 (we do this because if we are going to have a y-intercept then it is going to lay on the y-axis which means x would have to be 0)

to find x-intercept let y=0 (we do this because if we are going to have a x-intercept then it is going to lay on the x-axis which means y would have to be 0)

Answer 2

@myininaya do you know this one

Find the vertex of the following function. Enter your answer as an ordered pair, like this example: (-3, 12).
g(x) = - x2 + 6x – 8

Answer 3

Do you know the vertex form of a parabola?

Answer 4

kinda i dont understand this at all

Answer 5

\[y=a(x-h)^2+k \]
this is vertex form because it tells us the vertex is (h,k)
it also tells if the parabola is open up or down that is depending on value of a


so if you have
\[y=ax^2+bx+c \\ y=a(\frac{ax^2}{a}+\frac{bx}{a})+c \text{ I factored an a from the first two terms only } \\ y=a(x^2+\frac{b}{a}x)+c \text{ I need to complete the \square inside that parenthesis } \\ \text{ remember goal is to put in } y=a(x-h)^2+k \text{ form } \\ y=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2 \text{ I added \in (b/(2a))^2} \\ \text{ times that constant variable a so I also add \to subtract \it out } \\ y=a(x+\frac{b}{2a})^2+c-a \frac{b^2}{4a^2} \\ y=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a} \\ y=a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}\]

You should be able to compare this final equation to this
\[y=a(x-h)^2+k\]
to determine what are vertex is
when ever we have something in the form
\[y=ax^2+bx+c\]

Answer 6

looking at these two equations \[y=a(x-h)^2+k \text{ and } y=a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}\]
what would h have to be
and what would k have to be in order for these equations to be exactly the same?