How do you graph #f(x)=x^2+2x+5# and identify the x intercepts, vertex?
You can find the roots using the quadratic equation and the vertex using the formula
Recall the general form of a quadratic function:
VERTEX: The vertex has an
YINTERCEPT: The graph crosses the
XINTERCEPT: To find the
Substituting gives us
as our roots, or
Alternatively, we could try factorising the quadratic manually, this is easier if the quadratic has integer roots (or rational roots). In this case, we can't (well, we can, but not in terms of rational numbers).
Together, the roots, the vertex, and the
Also, since the leading coefficient is negative, you know that the quadratic must but an upsidedown U.
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To graph the function ( f(x) = x^2 + 2x + 5 ) and identify the xintercepts and vertex, follow these steps:

Plot the vertex: The vertex of the quadratic function ( f(x) = ax^2 + bx + c ) is given by the formula ( \left( \frac{b}{2a}, f\left(\frac{b}{2a}\right) \right) ). In this case, ( a = 1 ), ( b = 2 ), and ( c = 5 ). So, the xcoordinate of the vertex is ( \frac{2}{2(1)} = 1 ). Substitute ( x = 1 ) into the function to find the ycoordinate of the vertex: ( f(1) = (1)^2 + 2(1) + 5 = 6 ). Therefore, the vertex is ( (1, 6) ).

Find the xintercepts: The xintercepts are the points where the graph intersects the xaxis. To find them, set ( f(x) = 0 ) and solve for x. So, we need to solve the equation ( x^2 + 2x + 5 = 0 ) using the quadratic formula or factoring.
Using the quadratic formula: ( x = \frac{b \pm \sqrt{b^2  4ac}}{2a} ), where ( a = 1 ), ( b = 2 ), and ( c = 5 ). Substituting these values, we get ( x = \frac{2 \pm \sqrt{2^2  4(1)(5)}}{2(1)} ). Simplifying, ( x = \frac{2 \pm \sqrt{4 + 20}}{2} = \frac{2 \pm \sqrt{24}}{2} = \frac{2 \pm 2\sqrt{6}}{2} = 1 \pm \sqrt{6} ). So, the xintercepts are ( x = 1 + \sqrt{6} ) and ( x = 1  \sqrt{6} ).
 Plot the vertex and xintercepts on the graph and draw the parabola passing through these points.
So, the vertex of the function ( f(x) = x^2 + 2x + 5 ) is ( (1, 6) ), and the xintercepts are ( x = 1 + \sqrt{6} ) and ( x = 1  \sqrt{6} ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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