Question

Find the vertex, axis of symmetry, y intercept of each parabola. State the number of x intercepts. Then sketch the graph of each function. f(x)=2(x-4)^2+4

Answer 1

Answer:

Vertex: $(x,y) = (4,4)$, Axis of symmetry: $x = 4$, no x-Intercepts, y-Intercepts: $(x,y) = (0, 36)$. The graph is represented in the image attached below.

Step-by-step explanation:

The equation of the parabola in vertex form and whose axis of symmetry is vertical is described by this formula:

$y - k = C\cdot (x-h)^{2}$ (1)

Where:

$x$ - Independent variable.

$y$ - Dependent variable.

$C$ - Vertex constant.

$h, k$ - Coordinates of the vertex.

By direct comparison, we find the following information:

$h = 4$, $k = 4$, $C = 2$

Vertex

The vertex is a point of the parabola so that $(x,y) = (h, k)$.

If we know that $h = 4$ and $k = 4$, then the coordinates of the vertex are $(x,y) = (4,4)$.

Axis of symmetry

The axis of symmetry is a line of the form $x = h$.

If we know that $h = 4$, then the axis of symmetry is $x = 4$.

To find the x and y intercepts, we need to transform the equation of the parabola into its standards, which is a second grade polynomial:

$y - k = C\cdot (x-h)^{2}$

$y - k = C\cdot (x^{2}-2\cdot h\cdot x +h^{2})$

$y = C\cdot x^{2} -2\cdot C\cdot h \cdot x + (C\cdot h^{2}+k)$

If we know that $h = 4$, $k = 4$ and $C = 2$, then the equation of the parabola in standard form is:

$y = 2\cdot x^{2}-16\cdot x + 36$

x-Intercepts

The x-intercepts of the polynomial (if exist) can be found by the Quadratic Formula:

$x_{1,2} = \frac{16\pm \sqrt{(-16)^{2}-4\cdot 2\cdot (36)}}{4}$

$x_{1,2} = 4 \pm \frac{\sqrt{-32}}{4}$

$x_{1,2} = 4 \pm i \,\frac{4\sqrt{2}}{4}$

$x_{1,2} = 4 \pm i\,\sqrt{2}$

As both roots are conjugated complex numbers, there are no x-intercepts.

y-Intercepts

The y-intercept (if exists) can be found by evaluating the polynomial at $x = 0$:

$y = 2\cdot 0^{2} - 16\cdot 0 + 36$

$y = 36$

The y-intercept is $(x,y) = (0, 36)$.

Lastly, we proceed to plot the function by using graphing tools.