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svet-max [94.6K]
3 years ago
11

Graphing Quadratic Functions (help please )

Mathematics
1 answer:
Reil [10]3 years ago
6 0

Answer:

#1

Zeroes: x = 4, x = -6

The vertex is at ( -1 , 50 )

Step-by-step explanation:

I'll do the first one, you try the rest

All of these are written in y = a(x - x1)(x - x2) where x1 and x2 are given and x are the two different zeroes (or x-intercepts)

f(x) = -2(x-4)(x+6)

The zeroes are x - 4 = 0 and x + 6 = 0 if you solve, you get 4 and -6

Zeroes: x = 4, x = -6

The x value of the vertex can be found using -b/2a (in the form ax^2 + bx + c)

First expand f(x)

-2(x^2 + 2x - 24) => -2x^2 - 4x + 48

b = -4 and a = -2

-(-4)/2(-2) = 4/-4 = -1

the x-value of the vertex is -1

Now substitute -1 as x to find the y-value of the vertex

-2(-1-4)(-1+6) = -2(-5)(5) = 50 <= y-value of vertex

The vertex is at ( -1 , 50 )

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