Question

Can someone help me identify the vertex and x-intercepts of f(x)=-(x^2+x-30) and then check the results by writing the quadratic function in standard form??

Answer 1

there are a few ways to do this:
1) factor the quadratic to find the x-intercepts. the x value of the vertex will be exactly in between them. then use the equation to find the y value of the vertex

Answer 2

2.) complete the square, so that you can write the equation in "vertex form"

Answer 3

which way do you think would be better?

Answer 4

3) write the equation in standard form (i.e. distribute the -1) and then use
x value of vertex = -b/(2a)
where a,b (and c) are the coefficients y= ax^2 + bx+c

Answer 5

I would try to factor the quadratic.
list the factors of 30

Answer 6

I got (1,2,3,5,6,10,15,30)

Answer 7

1,30
2,15
3,10
5,6

are all of them
the minus sign on -30 (the last number) means the factors have different signs.
the plus sign on +1x means the larger one is positive.

you want the pair that gives you +1 when you add the pair (with the larger of the pair +, and the smaller of the pair -)

Answer 8

could it be +6 and -5?

Answer 9

yes, that means x^2+x-30
factors into (x-5)(x+6)
remember we have that minus sign out front: f(x)= -(x^2+x-30)
we get
f(x)= -(x-5)(x+6)

Answer 10

the x-intercepts make f(x) zero. in other words, solve for x in
x-5=0
and
x+6= 0

Answer 11

x=5 x=-6 but what about the negative sign out front?

Answer 12

the minus sign out front "flips" the parabola so it looks like a frown... but the x-intercepts are the same.

Answer 13

oh okay, so the x intercepts would still be 5 and -6?

Answer 14

yes. Here is a graph with and without the minus sign
the blue one is yours

Answer 15

what formula would i use to find the vertex?

Answer 16

the x value of the vertex is half-way between the x values of the x-intercepts

Answer 17

so its a number in between 5 and -6?

Answer 18

the number *exactly* in the middle between 5 and -6

Answer 19

one way to find the middle number is find the average of 5 and -6

Answer 20

-0.5?

Answer 21

yes, that is the x value of the vertex. to find the y value, use the equation
f(x)=-(x^2+x-30)

Answer 22

I got 29.25.. is that correct?

Answer 23

your calculator is broken

-( (-½)^2 + -½ -30)
-( ¼ - ½ -30)
-(-¼ -30)
-(-30.25)
30.25

Answer 24

oh i used .5 instead of -.5!

Answer 25

the vertex is at (-0.5,30.25)
the x-intercepts are at x= -6 and x=5

Answer 26

for this part
quadratic function in standard form??

I think we distribute the minus sign to get

f(x) = -x^2 -x +30

Answer 27

I don't know how that "checks" anything, unless they wanted you to write the equation in vertex form
y= a(x-h)^2 + k where (h,k) is the vertex
and then change it into standard form and show you get the same equation
btw, vertex form is
y= -(x - (-0.5))^2 + 30.25
y= -(x+0.5)^2 + 30.25

Answer 28

thank you so much, i really appreciate it!! :)