Question

a. Given the function y1 = x2 + 4x-12
Calculate -if applicable ─ intersections with the axes, minimum and maximum and the domain of the function.

Answer 1

\[x ^{2}\]

Answer 2

not x2

Answer 3

The domain are all "allowed" x values
usually every x is allowed unless it causes trouble (for example divide by zero, or square root of a negative number). For this equation, no x causes trouble, so the domain is
-infty to + infty, which you can write as
\( (-\infty , \infty) \)

Answer 4

For
intersections with the axes

for the y-axis, this happens when x=0. So put 0 in for x in the equation and simplify to get the y value where the curve intersects the y-axis

Answer 5

ahh so far i understand it now

Answer 6

what do you get for the intersection with the y-axis?

Answer 7

y=(0,-12)

Answer 8

intersection with the x-axis happens when y=0. you have to solve for x :
\[ x^2 + 4x-12=0 \]
this quadratic factors

Answer 9

can you factor it ?

Answer 10

for the max value, we can focus on the "highest order term", i.e. the x^2 term
as x gets large, x^2 is even larger. And because x can "go to infinity" , so does x^2
that means the max value is +infinity (or unbounded)

Answer 11

for the min value,
this curve is a parabola with a "smile" \( \cup\) shape
its smallest value is at its vertex, and the x value of the vertex is at
\[ x_{vertex}= \frac{-b}{2a} \]
where to find a and b, match your equation with
\[ y= a x^2 + bx +c \]

Answer 12

I do not know if i can factor it... i do not know how to solve that one.

Answer 13

Something tells me you already had that lesson.
But you do this:
look at the last number, the -12
list all pairs of numbers that when you multiply them, you get 12
they are:
1,12
2,6
3,4

Next, the - in front of the 12 means the factors will have different signs (one + the other -)
look at the number in front of the x, i.e. +4
the + means the larger number in each pair is positive. so put in the signs:
-1+12
-2+4
-3+4

now add each pair. if any pair adds up to +4 , those are the factors

Answer 14

**the middle pair should be -2+6

Answer 15

the middle pair adds up to +4

Answer 16

-2+6 = 4
so -2 and +6 go into
(x-2)(x+6)
if we multiply that out, we get x^2 +4x -12, which is what we started with

so the problem
\[ x^2 +4x -12=0 \]
can be written as
\[ (x-2)(x+6)=0 \]

now it is easy to see which x values make the left side zero.
any ideas?

Answer 17

2 and 6?

Answer 18

let's test 2:
(2-2)(2+6)
0*8 = 0 yes that worked

now 6:
(6-2)*(6+6)
4*12= 48. no, 6 does not work

Answer 19

-6?

Answer 20

-6 is much better , because -6+6 is 0

so the x-intercepts are at x=-6 and x=2

Answer 21

ok thanks :)

Answer 22

Can you find the min value?
You have to first find the x value of the vertex (see up above)

Answer 23

is it -24? i did \[-4\div2 = -2\] then used -2 in the formula

Answer 24

yes, x=-2

y= x^2 + 4x-12
y= (-2)*(-2) + 4*-2 -12
= 4 -8 -12
= -4-12
= -16

Answer 25

it looks like you did (-2)^2 and got -4, so try to figure out why you did the wrong.
-2 * -2 is +4

Answer 26

ahh yes I did not put -2 , sloppy error

Answer 27

thanks!