Identify the y-intercept and the axis of symmetry for the graph of f(x) = 10x^2 + 40x + 42.
a
42; x = 4
b
42; x = –2
c
–42; x = 2
d
0; x = –4

Answers

Answer 1

Answer:

b)42, x=-2

Step-by-step explanation:

Answer 2

Answer:

The option 'b' is correct.

Step-by-step explanation:

Given the function

[tex]f\left(x\right)\:=\:10x^2\:+\:40x\:+\:42[/tex]

Determining the y-intercept

The y-intercept can be obtained by setting the value x = 0

[tex]y\:=\:10x^2\:+\:40x\:+\:42[/tex]

[tex]y\:=\:10\left(0\right)^2\:+\:40\left(0\right)\:+\:42[/tex]

    [tex]= 0+0+42[/tex]

    [tex]=42[/tex]

Therefore, the y-intercept is:

(0, 42)

Determining the axis of symmetry

Given the equation

[tex]y=\:10x^2\:+\:40x\:+\:42[/tex]

For a parabola in standard form [tex]y=ax^2\:+\:bx\:+c[/tex]

the axis of symmetry is the vertical line that goes through the vertex

[tex]x=\frac{-b}{2a}[/tex]

so

The  axis of symmetry for [tex]y=ax^2\:+\:bx\:+c[/tex]  is [tex]x=\frac{-b}{2a}[/tex]

[tex]a=10,\:b=40[/tex]

[tex]x=\frac{-b}{2a}[/tex]

[tex]x=\frac{-40}{2\cdot \:10}[/tex]

[tex]x=-2[/tex]

Therefore, the option 'b' is correct.


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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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