Can you please help me with the these problems.
8.Write an equation in standard form of the parabola that has the same shape as the graph of f(x) = 11x^2, but which has its vertex at (2, 9).
10. Use the vertex and intercepts to sketch the graph of the quadratic function.
y + 3 = (x + 6)^2
8.
Clearly it must be y-9 = 11(x-2)^2
10.
sketch y = x^2, then move it 6 units to the left and 3 units up
5x-y=3
y=5x-3
Sure, I'd be happy to help you with these problems.
8. To write an equation in standard form of a parabola with a given vertex, we can use the formula:
f(x) = a(x - h)^2 + k
where (h, k) represents the vertex of the parabola. In this case, we are given that the vertex is (2, 9), so we can substitute these values into the formula:
f(x) = a(x - 2)^2 + 9
To find the value of 'a', we can use the fact that the shape of the parabola is the same as that of f(x) = 11x^2, which has a coefficient of 11. This means that the coefficient 'a' in our equation should also be 11.
So the equation in standard form of the parabola with the given vertex is:
f(x) = 11(x - 2)^2 + 9
10. To sketch the graph of the quadratic function y + 3 = (x + 6)^2, we can start by rearranging the equation to the standard form y = ax^2 + bx + c.
In this case, we can expand the equation:
y + 3 = (x + 6)^2
y = (x + 6)^2 - 3
y = x^2 + 12x + 36 - 3
y = x^2 + 12x + 33
Now, we can see that the coefficient of x^2 is 1, the coefficient of x is 12, and the constant term is 33.
To find the vertex of the parabola, we can use the formula x = -b / (2a). In this case, a = 1 and b = 12. So the x-coordinate of the vertex is:
x = -12 / (2*1) = -6
To find the y-coordinate of the vertex, we can substitute the x-coordinate into the equation:
y = (-6)^2 + 12(-6) + 33
y = 36 - 72 + 33
y = -3
So the vertex of the parabola is (-6, -3).
Next, we can find the x-intercepts by setting y = 0 and solving for x:
0 = x^2 + 12x + 33
This quadratic equation does not factor easily, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 12, and c = 33. Plugging these values into the formula, we get:
x = (-12 ± √(12^2 - 4*1*33)) / (2*1)
x = (-12 ± √(144 - 132)) / 2
x = (-12 ± √12) / 2
Simplifying further:
x = (-12 ± 2√3) / 2
x = -6 ± √3
So the x-intercepts are approximately -6 + √3 and -6 - √3.
Using the vertex and intercepts, we can sketch the graph of the quadratic function. The vertex is (-6, -3) and the x-intercepts are approximately -6 + √3 and -6 - √3. The graph will be a symmetric U-shape with the vertex at the lowest point on the graph. You can plot these points and connect them to get a rough sketch of the graph.