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How do you find the y-coordinate vertex of a parabola?
Once you calculate the X coordinate using the axis of symmetry (X=-b/2a), you plug that value in for all of the X's in the equation of the parabola. You then solve the equation for the value of Y.
How do you find the axis of symmetry?
X= -b / 2a
How do you find the x intercepts using the vertex and y intercept?
It depends on the vertex of what!
What do you know about a function that is made up of ordered pairs in such a way that the same y-value appears in a correspondence with two different x-values?
For example, y = ax2 + bx + c (the equation of a parabola). Every parabola has an axis of symmetry and the graph to either side of the axis of symmetry is a mirror image of the other side. It means that if we know a point on one side of the parabola, we can find its symmetric point to the other side, based on the axis of symmetry. Those symmetric points have opposite x-coordinate values, and the same y-coordinate value. The vertex only is a single point which lies on the axis of symmetry.
What is the process to graph quardratic equations?
If you are using a calculator just plug it in and hit graph. If you are doing it by hand, start with making a X-Y Table. Plug in X values into the equation to get a Y value out. Plot about 5 points on the graph to get a basic look at the parabola. To get the right the values, you want to start with the vertex and go out from there. To start, you need to find the axis of symmetry (-b/2a) [From the basic equation of ax squared +bx + c] That is the X Value for the vertex. Plug that in to find the Y Value for the vertex. The more points you find the more accurate the graph but normally 5 is enough (vertex and two on left and right)
Find the vertex and equation of the directri for y2 equals -32x?
y2 = 32x y = ±√32x the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
How do you find the axis of symmetry and vertex of y equals x squared plus 6x plus 10?
By completing the square y = (x+3)2+1 Axis of symmetry and vertex: x = -3 and (-3, 1) Note that the parabola has no x intercepts because the discriminant is less than zero
How do you find the y-coordinate vertex of a parabola?
Once you calculate the X coordinate using the axis of symmetry (X=-b/2a), you plug that value in for all of the X's in the equation of the parabola. You then solve the equation for the value of Y.
Find the equation of the axis symmetry and the coordinates of the vertex of the graph of each function for y equals 2x plus 4?
I'm assuming that you meant y = 2(x^2) +4. If it were only y = 2x +4, then this would be a linear equation and not a parabola. Anyways, use the equation x = -b/2a to find the x-value of your vertex AND your axis of symmetry. (Given the standard equation y = a(x^2) + bx + c) So, x = -0/2(2) - x = 0 (Axis of Symmetry) Now plug 0 back into your equation to find your y-value of your vertex. y = 2(0^2)+4 y=0 + 4 y = 4 Therefore Vertex = (0,4)
How do you find the axis of symmetry?
X= -b / 2a
Tell whether the graph opens up or down Find the coordinates of the vertex Write an equation of the axis of symmetry for the equation y equals 2x2 plus 3x plus 6?
y = 2x2 + 3x + 6 Since a > 0 (a = 2, b = 3, c = 6) the graph opens upward. The coordinates of the vertex are (-b/2a, f(-b/2a)) = (- 0.75, 4.875). The equation of the axis of symmetry is x = -0.75.
The equation for the axis of symmetry is?
Your equation must be in y=ax^2+bx+c form Then the equation is x= -b/2a That is how you find the axis of symmetry
How do you find the x intercepts using the vertex and y intercept?
It depends on the vertex of what!
What do you know about a function that is made up of ordered pairs in such a way that the same y-value appears in a correspondence with two different x-values?
For example, y = ax2 + bx + c (the equation of a parabola). Every parabola has an axis of symmetry and the graph to either side of the axis of symmetry is a mirror image of the other side. It means that if we know a point on one side of the parabola, we can find its symmetric point to the other side, based on the axis of symmetry. Those symmetric points have opposite x-coordinate values, and the same y-coordinate value. The vertex only is a single point which lies on the axis of symmetry.
What is the process to graph quardratic equations?
If you are using a calculator just plug it in and hit graph. If you are doing it by hand, start with making a X-Y Table. Plug in X values into the equation to get a Y value out. Plot about 5 points on the graph to get a basic look at the parabola. To get the right the values, you want to start with the vertex and go out from there. To start, you need to find the axis of symmetry (-b/2a) [From the basic equation of ax squared +bx + c] That is the X Value for the vertex. Plug that in to find the Y Value for the vertex. The more points you find the more accurate the graph but normally 5 is enough (vertex and two on left and right)
How do you find the gradient of a quadratic equation?
First the formula is g(x)=ax2+bx+c First find where the parabola cuts the x axis Then find the equation of the axis of symmetry Then
How do you find the equation of the axis of symmetry of y equals 2x plus 2 plus 4x plus 2?
y = 2x + 2 + 4x+ 2 = 6x + 4 This is NOT a symmetric function and so there is no axis of symmetry.
