Would the parabola described by -(x+2) +4 open upwards or downwards?
Q. I've been taught that a negative in front of everything in an equation like the above means that it opens downwards. However, it's squared, and a negative times a negative is a positive. So it would open upwards? My graphing calculator says it opens upwards, but I was skeptical. If it does open upwards, what is the equation for a parabola with a vertex at (-2,4) that opens downwards?
Asked by Donald J - Mon Mar 23 19:57:40 2009 - - 1 Answers - 0 Comments
Q. I've been taught that a negative in front of everything in an equation like the above means that it opens downwards. However, it's squared, and a negative times a negative is a positive. So it would open upwards? My graphing calculator says it opens upwards, but I was skeptical. If it does open upwards, what is the equation for a parabola with a vertex at (-2,4) that opens downwards?
Asked by Donald J - Mon Mar 23 19:57:40 2009 - - 1 Answers - 0 Comments
how do i know whether a parabola opens upwards or downwards?.?
Q. how do i know whether a parabola opens upwards or downwards?.?
Asked by fatkat - Thu Jun 5 00:40:03 2008 - - 8 Answers - 0 Comments
A. y=ax^2 + bx + c a>0 then up a<0 then down
Answered by mwcritt - Thu Jun 5 00:43:03 2008
Q. how do i know whether a parabola opens upwards or downwards?.?
Asked by fatkat - Thu Jun 5 00:40:03 2008 - - 8 Answers - 0 Comments
A. y=ax^2 + bx + c a>0 then up a<0 then down
Answered by mwcritt - Thu Jun 5 00:43:03 2008
Equation in intercept form for parabola?
Q. Write an equation in intercept form for a parabola which meets the x-axis at (4,0) and (-2,0). The parabola should open downwards like a frown. Help , please !
Asked by sO fReSh - Sat Sep 5 16:31:07 2009 - - 1 Answers - 0 Comments
A. Let f(x) be the quadratic function. Since the points (4,0) and (-2,0) are on the parabola, this tells us that (x-4) and (x--2)=(x+2) are factors of f(x). So f(x)=a(x+2)(x-4), where a is a real number. Since the parabola opens downward, then the coefficient of x^2 must be negative, i.e. a must be less than 0. So we can take a to be any negative number. Take a = -1. Then f(x) = -(x+2)(x-4). Expand and express in the desired form.
Answered by tangy - Sun Sep 6 02:03:54 2009
Q. Write an equation in intercept form for a parabola which meets the x-axis at (4,0) and (-2,0). The parabola should open downwards like a frown. Help , please !
Asked by sO fReSh - Sat Sep 5 16:31:07 2009 - - 1 Answers - 0 Comments
A. Let f(x) be the quadratic function. Since the points (4,0) and (-2,0) are on the parabola, this tells us that (x-4) and (x--2)=(x+2) are factors of f(x). So f(x)=a(x+2)(x-4), where a is a real number. Since the parabola opens downward, then the coefficient of x^2 must be negative, i.e. a must be less than 0. So we can take a to be any negative number. Take a = -1. Then f(x) = -(x+2)(x-4). Expand and express in the desired form.
Answered by tangy - Sun Sep 6 02:03:54 2009
Math Parabola questions! Thanks! Hint: y = a(x-h)^2 + k?
Q. 1. A vertical parabola opening downwards has a vertex of (0,11), x-intercepts of (3,0) and (-3,0). What is the standard form equation? 2. A vertical parabola opening downwards has a vertex of (2,11). The x-intercepts are (5,0) and (-1,0). What is the general form equation?
Asked by Rizr N - Mon Jun 8 19:50:46 2009 - - 1 Answers - 0 Comments
A. y = ax + 11 0 = 9a + 11 a = -11/9 y = (-11/9)x + 11 --- y = a(x-2) + 11 0 = a(5-2) + 11 a = -11/9 y = (-11/9)(x-2) + 11
Answered by unknown - Mon Jun 8 19:56:08 2009
Q. 1. A vertical parabola opening downwards has a vertex of (0,11), x-intercepts of (3,0) and (-3,0). What is the standard form equation? 2. A vertical parabola opening downwards has a vertex of (2,11). The x-intercepts are (5,0) and (-1,0). What is the general form equation?
Asked by Rizr N - Mon Jun 8 19:50:46 2009 - - 1 Answers - 0 Comments
A. y = ax + 11 0 = 9a + 11 a = -11/9 y = (-11/9)x + 11 --- y = a(x-2) + 11 0 = a(5-2) + 11 a = -11/9 y = (-11/9)(x-2) + 11
Answered by unknown - Mon Jun 8 19:56:08 2009
help, i need to know how to find the equation of a parabola?
Q. How do i find the equation for a parabola with a vertex of (-3,8) a axis of symmetry of x= -3, and 2 zeros of (-5,0) and (-1,0). I know one thing. The maximum is 8, and it opens downwards, so the coefficient of x squared should be negative. Please help, and thanks for your time.
Asked by REX - Sat May 3 23:44:37 2008 - - 1 Answers - 0 Comments
A. The vertex is (h, k) = (-3, 8). It has two zeros of (-5, 0) and (-1, 0). The line of symmetry is vertical so the parabola is vertical. It is therefore of the form: y = a(x + 5)(x + 1) = a(x + 6x + 5) Plug in the value for the vertex (-3, 8), and solve for a. 8 = a[(-3) + 6(-3) + 5] = a(9 - 18 + 5) = a(-4) a = -2 So the equation of the parabola is: y = -2(x + 6x + 5) y = -2x - 12x - 10
Answered by Northstar - Sat May 3 23:53:47 2008
Q. How do i find the equation for a parabola with a vertex of (-3,8) a axis of symmetry of x= -3, and 2 zeros of (-5,0) and (-1,0). I know one thing. The maximum is 8, and it opens downwards, so the coefficient of x squared should be negative. Please help, and thanks for your time.
Asked by REX - Sat May 3 23:44:37 2008 - - 1 Answers - 0 Comments
A. The vertex is (h, k) = (-3, 8). It has two zeros of (-5, 0) and (-1, 0). The line of symmetry is vertical so the parabola is vertical. It is therefore of the form: y = a(x + 5)(x + 1) = a(x + 6x + 5) Plug in the value for the vertex (-3, 8), and solve for a. 8 = a[(-3) + 6(-3) + 5] = a(9 - 18 + 5) = a(-4) a = -2 So the equation of the parabola is: y = -2(x + 6x + 5) y = -2x - 12x - 10
Answered by Northstar - Sat May 3 23:53:47 2008
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