Question

Find the X intercepts of the parabola with the vertex (1,-9) and y intercept of (0,-6)

Answer 1

y=a(x-h)^2+k  using the vetex (1,-9) for (h,k)

y=a(x-1)^2-9 and we are given the point (0,-6)

-6=a(-1)^2-9

-6=a-9

3=a

y=3(x-1)^2-9

The x-intercepts occur when y=0 so

3(x-1)^2-9=0  divide both sides by 3

(x-1)^2-3=0

(x-1)^2=3

x-1=±√3

x=1±√3

So the intercepts are the points:

(1+√3, 0) and (1-√3, 0)

Answer 2

The x-intercepts of the parabola with vertex (1,-9) and y-intercept of (0,-6) are of $x = 1 \pm \sqrt{3}$.

What is the equation of a parabola given it’s vertex?

The equation of a quadratic function, of vertex (h,k), is given by:

y = a(x - h)² + k

In which a is the leading coefficient.

In this problem, the parabola has vertex (1,-9), hence h = 1, k = -9, and:

y = a(x - 1)^2 - 9.

The y-intercept is of (0,-6), hence when x = 0, y = -6, and this is used to find a.

-6 = a - 9

a = 3.

So the equation is:

y = 3(x - 1)^2 - 9.

y = 3x² - 6x - 6.

The x-intercepts are the values of x for which:

3x² - 6x - 6 = 0.

Then:

x² - 2x - 2 = 0.

Which has coefficients a = 1, b = -2, c = -2, hence:

$\Delta = b^2 - 4ac = (-2)^2 - 4(1)(-2) = 12$

$x_1 = \frac{2 + \sqrt{12}}{2} = 1 + \sqrt{3}$

$x_2 = \frac{2 - \sqrt{12}}{2} = 1 - \sqrt{3}$

The x-intercepts of the parabola are $x = 1 \pm \sqrt{3}$.

More can be learned about quadratic equations at https://brainly.com/question/24737967

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