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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})

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", "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.

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$x = $ {x_vert}

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In part (b) use the formula that the axis of symmetry has equation $x = \\frac {-b}{2a}$. Rearranging and substituting a with -1.

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$b = 2 \\times ${x_vert} 

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$b =$ {b}

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In part (c) you now know the value of $b$ and have two coordinates to work with, so you can find $c$ by substitution.

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EITHER

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$0 = c +$ {b} $\\times${x_int1} $-(${x_int1}$)^2$

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$c = ($ {x_int1}$)^2$ $-${b}$\\times${x_int1}

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$c =$ {c}

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OR

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{y_vert} $= c +$ {b} $\\times${x_vert} $-${x_vert}$^2$

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$c =$ {x_vert}$^2$ $-${b}$\\times${x_vert} $+$ {y_vert}

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$c =$ {c}

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left intercept with the x-axis, must be negative to agree with diagram.

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x-coordinate of vertex, must be positive to agree with diagram.

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right intercept with the x-axis, calculated based on vertex

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y - coordinate of vertex, calculated from intercepts and leading coefficient of $x^2$ being -1

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coefficient of x

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constant term

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Find the equation of the axis of symmetry of $f$

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Find the value of $b$.

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Find the value of $c$.

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