How would you find the vertex and x-axis points of this equation so you could graph it? Please make it detailed and easy to understand. I don't really understand graphing :P
y>-2(x-5)(x+1)
y>-2(x-5)(x+1)
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You can simplify the equation by using the distributive property ( (x times x) + (x times 1) - (5times x) - (5)) Then multiply everythng by -2 and simplify.
The x axis points are 5 and -1, which plugged into inequality gives y>0, as x axis points are when y=0.
Y>-2(x^2 + x - 5x -5)
Y>-2(x^2 -4x -5)
Y> -2x^2 +8x + 10.
Use -b/2a to find x coordinate of vertex, and plus that value into equation to find y coordinate of vertex...so you get (2,8)............y=ax^2 + bx +c is the general form of quadratic equation....Thus mark (2,8) on the graph as the beginning of parabola, then mark the x axis points 5 and -2, and draw a general parabola shape that meets these points
The x axis points are 5 and -1, which plugged into inequality gives y>0, as x axis points are when y=0.
Y>-2(x^2 + x - 5x -5)
Y>-2(x^2 -4x -5)
Y> -2x^2 +8x + 10.
Use -b/2a to find x coordinate of vertex, and plus that value into equation to find y coordinate of vertex...so you get (2,8)............y=ax^2 + bx +c is the general form of quadratic equation....Thus mark (2,8) on the graph as the beginning of parabola, then mark the x axis points 5 and -2, and draw a general parabola shape that meets these points