SOLUTION: I forgot how to do this problem I am really struggling .. It says find the axis of symmetry, vertex, graph the function. And give the domain and range
F(x)=-x^2+10x-17
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Question 1004435: I forgot how to do this problem I am really struggling .. It says find the axis of symmetry, vertex, graph the function. And give the domain and range
F(x)=-x^2+10x-17
Found 2 solutions by Boreal, josgarithmetic:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
-x^2+10x+17=f(x)
vertex x value is -b/2a, which is -(-10/2)=5
f(5)=-25+50-17=8
vertex is at (5,8)
Symmetrical around x=5
Next step is y-intercept when x=0 and that is at (0,-17)
Because of symmetry, the change in x from 5 to 0 causes y to change from 8 to -17
That means from -5 to -10, going the other way, the y-value is the same, -17.
There are now 3 points, the vertex, and (0,-17) and (-10,-17). One can put in 1,2 or other values for x, but these are enough to graph it.
Domain is all x values.
Range are all y values less than or equal to 8, which is at the vertex.
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
Learn again how by studying this: http://www.algebra.com/my/Completing-the-Square-to-Solve-General-Quadratic-Equation.lesson?content_action=show_dev
Put into standard form and you can find roots (or zeros), the vertex, and if you let x=0 you find the y-axis intercept, helpful for graphing as well. The coefficient on leading term means the parabola has a maximum and the parabola opens downward.
Graph showing vertex and zeros,
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