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The graph of the quadratic function $f(x) = c + bx - x^2$ intersects the $x$-axis at the point A = ({x_int1},0) and has it's vertex at point B = ({x_vert},{y_vert})
\n", "advice": "In part (a) you can calculate the axis of symmetry from the x-coordinate of the vertex {x_vert}. Don't forget to put in the form of an equation for a vertical line.
\n$x = $ {x_vert}
\nIn part (b) use the formula that the axis of symmetry has equation $x = \\frac {-b}{2a}$. Rearranging and substituting a with -1.
\n$b = 2 \\times ${x_vert}
\n$b =$ {b}
\nIn part (c) you now know the value of $b$ and have two coordinates to work with, so you can find $c$ by substitution.
\nEITHER
\n$0 = c +$ {b} $\\times${x_int1} $-(${x_int1}$)^2$
\n$c = ($ {x_int1}$)^2$ $-${b}$\\times${x_int1}
\n$c =$ {c}
\nOR
\n{y_vert} $= c +$ {b} $\\times${x_vert} $-${x_vert}$^2$
\n$c =$ {x_vert}$^2$ $-${b}$\\times${x_vert} $+$ {y_vert}
\n$c =$ {c}
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