Express $-x^2+4x+12$ in the form $-(x+p)(x-q)$. Answer ... Sketch the graph of $y=-x^2+4x+12$ on the axes below. Indicate clearly the values where the graph crosses both axes. Find the coordinates of the turning point. Answer (..., ...) Without solving the equation algebraically, explain why $-x^2+4x+12=18$ has no solution. ...

1. Take out the common negative sign and factorise the quadratic inside the brackets.

$-x^2+4x+12=-(x^2-4x-12)$

Since $2\times(-6)=-12$ and $2+(-6)=-4$,

$x^2-4x-12=(x+2)(x-6)$.

Therefore,

$-x^2+4x+12=-(x+2)(x-6)$.

1. Find where the graph crosses the x-axis.

From part (a),
$y=-(x+2)(x-6).$
For x-intercepts, set $y=0$:
$-(x+2)(x-6)=0$
So
$x=-2 \quad \text{or} \quad x=6.$
Hence the graph crosses the x-axis at $(-2,0)$ and $(6,0)$.

2. Find where the graph crosses the y-axis and complete the sketch.

For the y-intercept, set $x=0$:
$y=-(0)^2+4(0)+12=12.$
So the graph crosses the y-axis at $(0,12)$.

Since the coefficient of $x^2$ is negative, the parabola opens downwards. Sketch a downward-opening parabola through $(-2,0)$, $(6,0)$ and $(0,12)$.

1. Find the x-coordinate of the turning point using the axis of symmetry.

From part (b), the x-intercepts are $x=-2$ and $x=6$.

The axis of symmetry is halfway between them:
$x=\frac{-2+6}{2}=2.$

2. Substitute the x-coordinate into the equation to find the y-coordinate.

When $x=2$,
$y=-(2)^2+4(2)+12=-4+8+12=16.$

So the turning point is $(2,16)$.

1. Compare the required value with the maximum value of the graph.

From part (c), the turning point is $(2,16)$. Since the parabola opens downwards, its maximum value is $16$.

The equation $-x^2+4x+12=18$ asks where the graph has $y=18$. But $18>16$, so the graph does not reach $y=18$. Hence there is no solution.