how to take x values in grap in polynomial
Answers
Answer:
To find the xxx-intercepts, we can solve the equation f(x)=0f(x)=0f, left parenthesis, x, right parenthesis, equals, 0.
\begin{aligned} f(x)&=(3x-2)(x+2)^2 \\\\ \tealD 0&= (3x-2)(x+2)^2\\ \\ \end{aligned}
f(x)
0
=(3x−2)(x+2)
2
=(3x−2)(x+2)
2
\begin{aligned}&\swarrow&\searrow\\\\ 3x-2&=0&\text{or}\quad x+2&=0&\small{\gray{\text{Zero product property}}}\\\\ x&=\dfrac{2}{3}&\text{or}\qquad x&=-2\end{aligned}
3x−2
x
↙
=0
=
3
2
↘
orx+2
orx
=0
=−2
Zero product property
The xxx-intercepts of the graph of y=f(x)y=f(x)y, equals, f, left parenthesis, x, right parenthesis are \left(\dfrac23,0\right)(
3
2
,0)left parenthesis, start fraction, 2, divided by, 3, end fraction, comma, 0, right parenthesis and (-2,0)(−2,0)left parenthesis, minus, 2, comma, 0, right parenthesis.
Our work also shows that \dfrac 23
3
2
start fraction, 2, divided by, 3, end fraction is a zero of multiplicity 111 and -2−2minus, 2 is a zero of multiplicity 222. This means that the graph will cross the xxx-axis at \left (\dfrac 23, 0\right)(
3
2
,0)left parenthesis, start fraction, 2, divided by, 3, end fraction, comma, 0, right parenthesis and touch the xxx-axis at (-2,0)(−2,0)left parenthesis, minus, 2, comma, 0, right parenthesis.
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Answer:
. PROCESS FOR determiners a polynomial
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