How do find the vertex and axis of symmetry, and intercepts for a quadratic equation #y=x^2 +2x -5#?
The vertex is at (
The axis of symmetry is
The
The
The standard form for the equation of a parabola is
Your equation is So Vertex Since The Insert this value of The vertex is at ( Axis of symmetry The axis of symmetry must pass through the vertex, so The axis of symmetry is To find the The To find the The Graph Now we prepare a chart. The axis of symmetry passes through Let's prepare a table with points that are 5 units on either side of the axis, that is, from
Plot these points.
And we have our graph.
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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 ):
- Vertex: ( \left( -\frac{2}{2}, \frac{4(1)(-5) - 2^2}{4(1)} \right) = (-1, -6) )
- Axis of symmetry: ( x = -\frac{2}{2} = -1 )
- x-intercepts: Set ( y = 0 ) and solve ( x^2 + 2x - 5 = 0 ) to get ( x = -3 ) and ( x = 1 ).
- y-intercept: Substitute ( x = 0 ) into the equation to get ( y = -5 ).
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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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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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