Here is an example of how to find the vertex of y=(x+1)2 + 2 by graphing.
First enter (x+1)2 + 2 into Y1 and graph it (press Y=, enter the equation, and then press GRAPH).
You should be able to see the vertex (if not press ZOOM, then ZStandard to reset back to the standard viewing window.) You need to know if the vertex is a minimum or a maximum. In this case, it is a minimum. Press 2ND CALC (above TRACE) and select minimum. (You would select maximum if the vertex were a maximum.
The calculator will ask for a left bound. Use the left arrow key to move the flashing start to the left of the vertex and then press ENTER. The calculator will then ask for a right bout. Move the flashing star to the right of the vertex and then press ENTER. The calculator will then ask for a guess. Move the flashing star close to the vertex (between the bounds you set) and pres ENTER.
Then calculator will then display the coordinates of the vertex. In this case, it will say that x is -1 and y is 2.
To determine the coordinates of the preimage of vertex M, I would need additional information about the transformation that was applied to vertex M, such as the type of transformation (e.g., translation, rotation, reflection, scaling) and the coordinates of M itself. If you provide the coordinates of M and the details of the transformation, I can help you find the preimage coordinates.
guess where it would go... shouldnt be cheating on your study island:)
2x2 + 4 + 1 = 2x2 + 5 So, the vertex is (0, 5)
To determine the coordinates of the vertex of a quadratic function in the form (y = ax^2 + bx + c), you can use the vertex formula (x = -\frac{b}{2a}) to find the x-coordinate. Once you have the x-coordinate, substitute it back into the original equation to find the corresponding y-coordinate. Thus, the vertex coordinates are ((-\frac{b}{2a}, f(-\frac{b}{2a}))). For a parabola, this point represents either the maximum or minimum value, depending on the sign of (a).
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 vertex of a triangle is the point where two or more sides of the triangle intersect. In the case of triangle TIF, the vertex would be the point where the sides TI and IF intersect. To determine the exact coordinates of the vertex, you would need the coordinates of points T, I, and F and then use the equations of the lines containing the sides to find their point of intersection.
Use this form: y= a(x-h)² + k ; plug in the x and y coordinates of the vertex into (h,k) and then the other point coordinates into (x,y) and solve for a.
To determine the coordinates of the fourth vertex of a rectangle, you need to know the coordinates of the other three vertices. If you have the coordinates of three vertices, you can find the fourth by using the properties of a rectangle, where opposite sides are equal and the diagonals bisect each other. For example, if the vertices are A(x1, y1), B(x2, y2), and C(x3, y3), you can find the fourth vertex D(x4, y4) through the midpoint formula or by ensuring that the lengths of the sides and the diagonals are consistent. Please provide the coordinates of the existing vertices for a specific answer.
To find the vertex of the parabola given by the equation ( y = x^2 + 2x - 3 ), we can use the vertex formula ( x = -\frac{b}{2a} ). Here, ( a = 1 ) and ( b = 2 ), so ( x = -\frac{2}{2 \cdot 1} = -1 ). Substituting ( x = -1 ) back into the equation gives ( y = (-1)^2 + 2(-1) - 3 = -4 ). Therefore, the coordinates of the vertex are ( (-1, -4) ).
-2-5
Find the midpoint of the two diagonals
Use the equation: (Y-k)^2 = 4a(X-h)