1. **Stating the problem:** We have four quadratic functions with parameter $m > 6$, $m \in \mathbb{N}$: $$f(x) = x^2 - 6x + m$$ $$g(x) = (x - 6)(x + m)$$ $$h(x) = (x - 6)^2 + m$$ $$j(x) = -(x + m)(x - 6)$$ We need to match each function to one of the four given graphs numbered 1 to 4 based on their properties. 2. **Analyzing each function:** - **Function $f(x)$:** - Standard form: $f(x) = x^2 - 6x + m$ - Since the coefficient of $x^2$ is positive, the parabola opens upward. - Vertex: Use vertex formula $x = -\frac{b}{2a} = \frac{6}{2} = 3$. - Vertex $y$-value: $f(3) = 3^2 - 6 \times 3 + m = 9 - 18 + m = m - 9$. - Since $m > 6$, $m - 9$ can be positive or negative depending on $m$. - **Function $g(x)$:** - Factored form: $g(x) = (x - 6)(x + m)$ - Roots at $x=6$ and $x=-m$. - Since $m > 6$, $-m < -6$, so roots are negative and positive. - Leading coefficient is positive, so parabola opens upward. - Vertex is between roots, so vertex $x$-coordinate is $\frac{6 + (-m)}{2} = \frac{6 - m}{2} < 0$ (since $m > 6$), so vertex is left of y-axis. - **Function $h(x)$:** - Vertex form: $h(x) = (x - 6)^2 + m$ - Parabola opens upward (coefficient of squared term positive). - Vertex at $(6, m)$. - Since $m > 6$, vertex is above x-axis and to the right of y-axis. - **Function $j(x)$:** - Factored form: $j(x) = -(x + m)(x - 6)$ - Roots at $x = -m$ and $x = 6$. - Leading coefficient negative, so parabola opens downward. - Vertex $x$-coordinate is midpoint of roots: $\frac{-m + 6}{2} < 0$ (since $m > 6$), so vertex is left of y-axis. 3. **Matching graphs:** - Graph 1: Upward-opening parabola with vertex labeled P, above x-axis and to the right of y-axis. - Matches $h(x)$ (vertex at $(6,m)$, $m>6$). - Graph 2: Downward-opening parabola with vertex on y-axis above x-axis, left x-intercept labeled Q, right branch crossing x-axis right of y-axis. - Vertex on y-axis means vertex $x=0$. - $j(x)$ vertex $x$-coordinate is $\frac{-m+6}{2}$ which is negative, not zero. - Check $f(x)$ vertex $x=3$, no. - Check $g(x)$ vertex $x=\frac{6 - m}{2}$ negative, no. - None have vertex exactly on y-axis, but $j(x)$ is downward opening and roots at $-m$ and $6$. - Since vertex is not on y-axis for $j(x)$, but graph 2 vertex is on y-axis, this suggests $j(x)$ is graph 2 with vertex shifted slightly left. - Graph 3: Upward-opening parabola with vertex below x-axis, crossing x-axis at two points, right x-intercept labeled R. - $f(x)$ vertex $y$-value is $m-9$. - For $m=7$, $f(3) = -2$ (below x-axis), so vertex below x-axis. - $f(x)$ opens upward and crosses x-axis twice. - So $f(x)$ matches graph 3. - Graph 4: Upward-opening parabola with vertex labeled S on y-axis below x-axis, crossing x-axis at two points. - $g(x)$ vertex $x$-coordinate is $\frac{6 - m}{2}$, negative. - But vertex on y-axis means $x=0$. - None of the functions have vertex exactly on y-axis except possibly $g(x)$ if $m=6$ (not allowed since $m>6$). - $g(x)$ opens upward, roots at $-m$ and $6$. - So graph 4 matches $g(x)$. 4. **Final matching:** - $f(x)$: graph 3 - $g(x)$: graph 4 - $h(x)$: graph 1 - $j(x)$: graph 2 **Answer:** 3142