Remember that the vertex form of a parabola or quadratic equation is:
y=a(x-h)^2+k, where (h,k) is the "vertex" which is the maximum or minimum point of the parabola (and a is half the acceleration of the of the function, but that is maybe too much :P)
In this case we are given that the vertex is (1,1) so we have:
y=a(x-1)^2+1, and then we are told that there is a point (0,-3) so we can say:
-3=a(0-1)^2+1
-3=a+1
-4=a so our complete equation in vertex form is:
y=-4(x-1)^2+1
Now you wish to know where the x-intercepts are. x-intercepts are when the graph touches the x-axis, ie, when y=0 so
0=-4(x-1)^2+1 add 4(x-1)^2 to both sides
4(x-1)^2=1 divide both sides by 4
(x-1)^2=1/4 take the square root of both sides
x-1=±√(1/4) which is equal to
x-1=±1/2 add 1 to both sides
x=1±1/2
So x=0.5 and 1.5, thus the x-intercept points are:
(0.5, 0) and (1.5, 0) or if you like fractions:
(1/2, 0) and (3/2, 0) :P
<span>you can solve all quadratic equations using this method:
---
x^2 + 4x + 4 = 0
---
the above quadratic equation is in standard form, with a=1, b=4, and c=4
---
to solve the quadratic equation, by using the quadratic formula, copy and paste this:
1 4 4
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
---
this quadratic has ONE real root (the x-intercept), which is:
x = -2
x = -2
---
---
---
---
x^2 + x = 0
---
the above quadratic equation is in standard form, with a=1, b=1, and c=0</span>
Answer:
2
Step-by-step explanation:
we start simplifying by removing the parenthesis
Multiply the exponents inside the the parenthesis
3^4 * 2^4
Now we apply exponential property
a^m * a^n = a^ (m+n)
3^4 * 3^-3 = 3^ (4-3) = 3^1
3 or 3^1 are same
3^1 at the top and bottom so we cancel it out
\frac{2^4}{2^3}
we apply log property . a^m / a^n = a^m-n
Now subtract the exponents
2^(4-3) = 2^1 = 2
Answer:
Are there directions?
Step-by-step explanation:
