How do you graph #f(x)=-x^2+2x+5# and identify the x intercepts, vertex?

Answer 1

You can find the roots using the quadratic equation and the vertex using the formula #-b/{2a}#.

Recall the general form of a quadratic function: #f(x)=ax^2+bx+c#.

VERTEX: The vertex has an #x#-coordinate of #-b/{2a}# (you can figure this out with calculus, or by factorising the general expression into turning point form). Once we know the #x#-coordinate, we can substitute into #f(x)# to find the #y#-coordinate. In this case, we end up with the coordinates #(1, 6)# for the vertex.

Y-INTERCEPT: The graph crosses the #y#-axis when #x=0#, so substitute #x=0# and we find #y=f(0)=c#. Thus, for a quadratic of the form #f(x)=ax^2+bx+c#, the #y#-intercept is always c. In this case, the #y#-intercept is 5.

X-INTERCEPT: To find the #x#-intercept, we have to solve the equation #ax^2+bx+c=0#. This has to be done through factorisation. By factorising the general quadratic form (Google it), we get the quadratic equation:

#x={-b+-sqrt(b^2-4ac)}/{2a}#

Substituting gives us

#x=1+-\sqrt(6)#

as our roots, or #x#-intercepts.

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 #y#-intercept give you 4-points. Which is more than enough to graph the quadratic freehand. This is what it looks like:

Also, since the leading coefficient is negative, you know that the quadratic must but an upside-down U.

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Answer 2

To graph the function ( f(x) = -x^2 + 2x + 5 ) and identify the x-intercepts and vertex, follow these steps:

  1. 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 x-coordinate of the vertex is ( \frac{-2}{2(-1)} = 1 ). Substitute ( x = 1 ) into the function to find the y-coordinate of the vertex: ( f(1) = -(1)^2 + 2(1) + 5 = 6 ). Therefore, the vertex is ( (1, 6) ).

  2. Find the x-intercepts: The x-intercepts are the points where the graph intersects the x-axis. 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 x-intercepts are ( x = 1 + \sqrt{6} ) and ( x = 1 - \sqrt{6} ).

  1. Plot the vertex and x-intercepts 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 x-intercepts are ( x = 1 + \sqrt{6} ) and ( x = 1 - \sqrt{6} ).

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Answer from HIX Tutor

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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