SOLUTION: {{{f(x)=25-x^2}}}The instructions say to sketch the graph of the quadratic function. Identify the vertex & x-intercept(s). I seem to be lost. I know to 1st factor it (I don't belie
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Question 53308This question is from textbook college algebra a graphing approach
: The instructions say to sketch the graph of the quadratic function. Identify the vertex & x-intercept(s). I seem to be lost. I know to 1st factor it (I don't believe this can be factored). Then complete the square. I think this gives me x=2.549 & x-2.549 so,x={2.549,2.549} I do believe something went very wrong here & beyond. HELP!
Thanks Bunches
This question is from textbook college algebra a graphing approach
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
This is the difference between two perfect squares.
In your case a=5 and b=x
Let f(x)=0
Use the zero product property and set each parenthesis = 0.
5+x=0
-5+5+x=0-5
x=-5
and
5-x=0
-5-x=0-5
-x=-5
-x/-1=-5/-1
x=5
Those are your x intersepts:(5,0),(-5,0).
Let x=0 and find your y intersept:
y=25-0^2
y=25
Now you can plot your y-intersept:(0,25)
This also happens to be the vertex of the parabola in this case.
The fact that x^2 is negative tells you that the parabola opens upside down.
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