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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
What is the axis of symmetry for y-x2 2x-4?
To find the axis of symmetry for the quadratic equation ( y = -x^2 + 2x - 4 ), you can use the formula ( x = -\frac{b}{2a} ), where ( a ) and ( b ) are the coefficients from the equation in standard form ( y = ax^2 + bx + c ). Here, ( a = -1 ) and ( b = 2 ). Plugging in the values, the axis of symmetry is ( x = -\frac{2}{2 \times -1} = 1 ). Thus, the axis of symmetry is ( x = 1 ).
How do you find the axis of symmetry?
X= -b / 2a
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 u find the equation of the axis of symmetry and the vertex of the graph of each function for example y x2-8x-9 Plz help i need to know this?
To find the equation of the axis of symmetry for the quadratic function (y = x^2 - 8x - 9), use the formula (x = -\frac{b}{2a}), where (a = 1) and (b = -8). This gives (x = -\frac{-8}{2 \cdot 1} = 4). The vertex can be found by substituting this (x) value back into the original equation: (y = 4^2 - 8(4) - 9 = 16 - 32 - 9 = -25). Thus, the vertex is at the point ((4, -25)) and the axis of symmetry is the line (x = 4).
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
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 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
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.
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)
What is the axis of symmetry for y-x2 2x-4?
To find the axis of symmetry for the quadratic equation ( y = -x^2 + 2x - 4 ), you can use the formula ( x = -\frac{b}{2a} ), where ( a ) and ( b ) are the coefficients from the equation in standard form ( y = ax^2 + bx + c ). Here, ( a = -1 ) and ( b = 2 ). Plugging in the values, the axis of symmetry is ( x = -\frac{2}{2 \times -1} = 1 ). Thus, the axis of symmetry is ( x = 1 ).
How do you find the axis of symmetry for a quadratic equation?
Complete the square, then find the value of x that would make the bracket zero ax^2 + bx + c = 0 line of symmetry is x = (-b/2a)
How do you find the axis of symmetry?
X= -b / 2a
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 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 equation of the line given the slope equals 5 and the x intercept equals 7?
Call the intercept with the y-axis b. The equation for the line is then y=5x+b With the information about the x-intercept we get: 0=5*7+b b=-35
How do you find the axis of symmetry of the quadratic function.?
The axis of symmetry of a quadratic function in the form (y = ax^2 + bx + c) can be found using the formula (x = -\frac{b}{2a}). This vertical line divides the parabola into two mirror-image halves. To find the corresponding (y)-coordinate, substitute the axis of symmetry value back into the quadratic function.
