Find the vertex form of the function. Then find each of the following. (A) Intercepts (B) Vertex (C) Maximum or minimum (D) Range s(x) = -5x^2 - 30x - 42 s(x) = (Answer has to be given in vertex form).

Janiya Rose

Janiya Rose

Answered question

2022-08-12

Find the vertex form of the function. Then find each of the following.
(A) Intercepts (B) Vertex (C) Maximum or minimum (D) Range
s ( x ) = 5 x 2 30 x 42
s(x) = (Answer has to be given in vertex form).

Answer & Explanation

Jaxson White

Jaxson White

Beginner2022-08-13Added 15 answers

The equation of a quadratic in vertex form is:
y = a ( x h ) 2 + k .
The vertex is at the point (h, k).
You must complete the square for this.
s ( x ) = 5 x 2 30 x 42
Take out a -5 to get:
5 ( x 2 + 6 x + ( 42 5 ) )
Complete the square :
s ( x ) = 5 ( x 2 + 6 x + ( 6 / 3 ) 2 + ( 42 / 5 ) ( 6 / 3 ) 2 ) s ( x ) = 5 ( x 2 + 6 x + 9 + ( 42 / 5 ) 9 ) s ( x ) = 5 [ ( x + 3 ) 2 ( 42 / 5 ) ( 45 / 5 ) ] s ( x ) = 5 [ ( x + 3 ) 2 ( 3 / 5 ) ] s ( x ) = 5 ( x + 3 ) 2 + 3.
The y-intercept is when x = 0, it's:
y = 5 ( 3 ) 2 + 3 = 45 + 3 = 42.
y-intercept is y = -42.
The x-intercept(s) are when y = 0, i.e. when s(x) = 0.
0 = 5 ( x + 3 ) 2 + 3
3 = 5 ( x + 3 ) 2
3 / 5 = ( x + 3 ) 2
x + 3 = + ( 3 5 )
x = 3 + ( 3 5 )
The two x-intercepts are: x = 3 + ( 3 5 ) \ \ x = 3 ( 3 5 )
vertex is at : (-3, 3).
There is a max since the parabola is upside down. The max is the same as the vertex, and it's at
x = 3.
Range is from negative infinity to the vertex.
( , 3 ]

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