Question

consider the function (h(x)=\frac{1}{2}x^{2}-4x + 9). what is the vertex of (h)? what is the equation of the line of symmetry of (h)? (h) has a select an answer of. the (x)-intercept(s) of (h) is/are. the (y)-intercept of (h) is.

Explanation:

Step1: Recall vertex - formula for a quadratic function

The quadratic function is in the form $h(x)=ax^{2}+bx + c$, where for $h(x)=\frac{1}{2}x^{2}-4x + 9$, $a=\frac{1}{2}$, $b=-4$, $c = 9$. The x - coordinate of the vertex is $x=-\frac{b}{2a}$.
$x=-\frac{-4}{2\times\frac{1}{2}}=\frac{4}{1}=4$.

Step2: Find the y - coordinate of the vertex

Substitute $x = 4$ into $h(x)$: $h(4)=\frac{1}{2}(4)^{2}-4\times4 + 9=\frac{1}{2}\times16-16 + 9=8-16 + 9=1$. So the vertex is $(4,1)$.

Step3: Find the equation of the line of symmetry

The line of symmetry of a quadratic function $y = ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. Since $x = 4$ is the x - coordinate of the vertex, the equation of the line of symmetry is $x = 4$.

Step4: Determine the minimum or maximum

Since $a=\frac{1}{2}>0$, the parabola opens upward and the function has a minimum value. The minimum value is $y = 1$ (the y - coordinate of the vertex).

Step5: Find the x - intercepts

Set $h(x)=0$, so $\frac{1}{2}x^{2}-4x + 9=0$. Multiply through by 2 to get $x^{2}-8x + 18=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b=-8$, $c = 18$. Then $\Delta=b^{2}-4ac=(-8)^{2}-4\times1\times18=64 - 72=-8<0$. So there are no real x - intercepts.

Step6: Find the y - intercept

Set $x = 0$ in $h(x)$. Then $h(0)=\frac{1}{2}(0)^{2}-4(0)+9 = 9$.

Answer:

Vertex: $(4,1)$
Line of symmetry: $x = 4$
Minimum value of 1 (since $a>0$)
x - intercepts: None
y - intercept: 9