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    how would i sketch the graph of y=3x^2+5x?? with all the points where it intersects with the axis?
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    (Original post by Chelsea12345)
    how would i sketch the graph of y=3x^2+5x?? with all the points where it intersects with the axis?
    You can factor x out to give x(3x+5)=(x)(3x+5), so crossing at (0,0) and (-5/3,0)
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    well to find where a graph intercepts the axis, you need to equate y to 0 and then equate x to 0, and go from there

    as it is a quadratic, you know it can either be one of two shapes depending on whether the coefficient of x^2 is negative or not, so from that you can sketch its shape
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    (Original post by Aear)
    You can factor x out to give x(3x+5)=(x)(3x+5), so crossing at (0,0) and (-5/3,0)
    Thankyou!! I was struggling to find a way for so long!!
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    (Original post by Chelsea12345)
    how would i sketch the graph of y=3x^2+5x?? with all the points where it intersects with the axis?
    Find the stationary point, recognise that it is a parabola, find the roots and y-intercept, and sketch it.
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    factorise it to work out the roots
    complete the square to calculate the minimum point.
    the y intercept is the value of 'c' in the equation in the form y=ax^2+bx+c (in ur case, ur intercept would be 0)
 
 
 
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