Question

cc.2.2.hs.d.10
identify the y-intercept and the axis of symmetry for the graph of f(x) = 10x² + 40x + 42.

○ 42; x = 4
○ 0; x = -4
○ 42; x = -2
○ -42; x = 2

Explanation:

Step1: Find the y-intercept

The y-intercept of a function \( f(x) \) is the value of \( f(0) \). For \( f(x)=10x^{2}+40x + 42 \), substitute \( x = 0 \):
\( f(0)=10(0)^{2}+40(0)+42=42 \). So the y - intercept is 42.

Step2: Find the axis of symmetry

For a quadratic function in the form \( f(x)=ax^{2}+bx + c \), the axis of symmetry is given by the formula \( x=-\frac{b}{2a} \).
Here, \( a = 10 \) and \( b = 40 \). Substitute these values into the formula:
\( x=-\frac{40}{2\times10}=-\frac{40}{20}=- 2 \). So the axis of symmetry is \( x=-2 \).

Answer:

\( 42; x = - 2 \) (corresponding to the option "42; \( x=-2 \)")