For questions 1 and 2, use factoring to find the x-intercepts of these quadratic equations.

1 answer:

6 0

1.x^2 +6x - 4 = 6x
X^2 + 6x -4 - 6x = 0
X^2 - 4 = 0
By difference of two squares :
X^2 - 4 = 0 can be written as (x-2) (x+2) = 0
(X-2)=0 therefore x= 2

(X+2)=0 therefore x= -2

2. X^2 - 8x = -6x
X^2 -8x + 6x = 0
X^2 - 2x = 0
X(X-2) = 0
X= 0, X= 2

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Someone do this got me rq

skelet666 [1.2K]

Answer:

The last listed functional expression:

$\left \{ {x+1\,\,\,\,{x\geq 2} \atop {x+2 \,\,\,\,x$

Step-by-step explanation:

It is important to notice that the two linear expressions that render such graph are parallel lines (same slope), and that the one valid for the left part of the domain, crosses the y-axis at the point (0,2), that is y = 2 when x = 0. On the other hand, if you prolong the line that describes the right hand side of the domain, that line will cross the y axis at a lower position than the previous one (0,1), that is y=1 when x = 0. This info gives us what the y-intercepts of the equations should be (the constant number that adds to the term in x in the equations: in the left section of the graph, the equation should have "x+2", while for the right section of the graph, the equation should have x+1.

It is also important to understand that the "solid" dot that is located in the region where the domain changes, (x=2) belongs to the domain on the right hand side of the graph, So, we are looking for a function definition that contains $x+1$ for the function, for the domain: $x\geq 2$ .