Question

1. Identify the y-intercept and the axis of symmetry for the graph of

f(x) = 10x2 + 40x + 42.
A. 42; x = 4 B. 0; x = -4 C. 42; x = -2 D. – 42; x = 2

Answer 1

The correct option is C). 42,x=(-2)

Point (0,42) is y-intercept of f(x)

x+2=0 is equation of axis of symmetry

Explanation:

The given function is f(x)=
`10x^(2) +40x+42`

To find y-intercept:

The y-intercept is point were x=0

y=f(x)=
`10x^(2) +40x+42`

y=f(0)=
`10(0)^(2) +40(0)+42`

y=42

Therefore,Point (0,42) is y-intercept of f(x)

To find axis of symmetry of f(x):

We know that f(x) is equation of parabola

Hence, the x -coordinate of the vertex of the parabola is the equation of the axis of symmetry.

if the equation of parabola is f(x)=
`ax^(2) +bx+c` then, the axis of symmetry is given by x=
`-(b)/(2a)`

Therefore, for f(x)=
`10x^(2) +40x+42`

The axis of symmetry will be,

x=
`-(b)/(2a)`

a=10 and b=40

x=
`-(40)/(2(10))`

x=-2

x+2=0 is equation of axis of symmetry

Thus,

The correct option is C). 42,x=(-2)

Note: In figure, blue curve is f(x) and black line is axis of symmetry