How do find the vertex and axis of symmetry, and intercepts for a quadratic equation #y=x^2 +2x -5#?

Answer 1

The vertex is at (#-1,-2#).
The axis of symmetry is #x = 1#.
The #y#-intercept is at (#0,5#).
The #x#-intercepts are at #x = -1- sqrt6# and #x = -1 + sqrt 6#.

The standard form for the equation of a parabola is

#y= a^2 + bx +c#

Your equation is

#y = x^2 + 2x - 5#

So

#a = 1#, #b = 2#, and #c = 5#.

Vertex

Since #a > 0#, the parabola opens upwards.

The #x#-coordinate of the vertex is at

#x = –b/(2a) = -2/(2×1) = -2/2 = -1#.

Insert this value of #x# back into the equation.

#y = x^2 + 2x - 5 = (-1)^2 + 2(-1) - 5 = 1 - 2 - 5 = -6#

The vertex is at (#-1,-2#).

Axis of symmetry

The axis of symmetry must pass through the vertex, so

The axis of symmetry is #x = -1#.

#y#-intercept

To find the #y#-intercept, we set #x = 0# and solve for #y#.

#y = x^2 + 2x – 5 = 0^2 + 2×0 -5 = 0 + 0 – 5 = -5#

The #y#-intercept is at (#0,5#).

#x#-intercepts

To find the #x#-intercepts, we set #y = 0# and solve for #x#.

#y = x^2 + 2x – 5#

#0 = x^2 + 2x – 5#

#x = (-b ±sqrt(b^2-4ac))/(2a) #

#x = (-2 ± sqrt(2^2 - 4×1×(-5)))/(2×1) #

#x = (-2 ± sqrt(2^2 - 4×1×(-5)))/(2×1)#

#x = (-2 ± sqrt(4 + 20))/2#

#x = (-2 ± sqrt24)/2#

#x = -1 ± 1/2sqrt24#

#x = -1 ± 1/2(sqrt(4×6))#

#x = -1 ± 1/2(2sqrt6)#

#x = -1 ± sqrt6#

The #x#-intercepts are at #x = -1- sqrt6# and #x = -1 + sqrt 6#.

Graph

Now we prepare a chart.

The axis of symmetry passes through #x = -1#.

Let's prepare a table with points that are 5 units on either side of the axis, that is, from #x = -6# to #x = 4#.

Plot these points.

And we have our graph.

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Answer 2

To find the vertex of a quadratic equation in the form ( y = ax^2 + bx + c ), where ( a \neq 0 ), the vertex coordinates are given by ( \left( -\frac{b}{2a}, \frac{4ac - b^2}{4a} \right) ). The axis of symmetry is the vertical line passing through the vertex, given by ( x = -\frac{b}{2a} ). To find the x-intercepts, set ( y = 0 ) and solve for ( x ). To find the y-intercept, substitute ( x = 0 ) into the equation.

For the given equation ( y = x^2 + 2x - 5 ):

  1. Vertex: ( \left( -\frac{2}{2}, \frac{4(1)(-5) - 2^2}{4(1)} \right) = (-1, -6) )
  2. Axis of symmetry: ( x = -\frac{2}{2} = -1 )
  3. x-intercepts: Set ( y = 0 ) and solve ( x^2 + 2x - 5 = 0 ) to get ( x = -3 ) and ( x = 1 ).
  4. y-intercept: Substitute ( x = 0 ) into the equation to get ( y = -5 ).
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Answer 3

To find the vertex of the quadratic equation (y = x^2 + 2x - 5), we first need to determine the axis of symmetry, which is given by (x = -\frac{b}{2a}) where (a = 1) (coefficient of (x^2)) and (b = 2) (coefficient of (x)). So, (x = -\frac{2}{2} = -1).

To find the corresponding (y)-coordinate of the vertex, substitute (x = -1) into the equation: [y = (-1)^2 + 2(-1) - 5 = -4.]

So, the vertex is at ((-1, -4)).

The axis of symmetry is the vertical line passing through the vertex, which is (x = -1).

To find the (x)-intercepts, set (y = 0) and solve for (x): [x^2 + 2x - 5 = 0.]

Using the quadratic formula: [x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}.]

Plugging in (a = 1), (b = 2), and (c = -5), we get: [x = \frac{{-2 \pm \sqrt{{2^2 - 4(1)(-5)}}}}{{2(1)}} = \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}).

To find the (y)-intercept, set (x = 0): [y = (0)^2 + 2(0) - 5 = -5.]

So, the (y)-intercept is at ((0, -5)).

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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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