How do you find the vertex and the intercepts for # f(x)= x^2+10x-8#?
Vertex is at
y intercept is at
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To find the vertex and intercepts for ( f(x) = x^2 + 10x - 8 ):
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Vertex: Use the formula for the x-coordinate of the vertex: ( x = \frac{-b}{2a} ), where ( a = 1 ) and ( b = 10 ). Plug these values into the formula to find the x-coordinate of the vertex. Then, substitute this value back into the original function to find the corresponding y-coordinate.
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x-intercepts (zeros): Set ( f(x) ) equal to zero and solve for ( x ). These are the points where the graph crosses the x-axis.
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y-intercept: Set ( x = 0 ) in the function and solve for ( y ). This gives the point where the graph crosses the y-axis.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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