In the event that a line has no y-intercept, that implies it never converges the y-intercept, so it must be parallel to the y-intercept. This implies it is a vertical line, for example, . This slant of this line is vague. In the event that the line has no x-intercept, at that point it never meets the x-intercept, so it must be parallel to the x-pivot.
If x-intercept is -2 the line would be vertical as the y-intercept = 0.
Answer:
![-\frac{3\sqrt[3]{t} }{2}](https://tex.z-dn.net/?f=-%5Cfrac%7B3%5Csqrt%5B3%5D%7Bt%7D%20%7D%7B2%7D)
Step-by-step explanation:
1: Write g(t) as y, resulting in 
2: Interchange the variables y and t, resulting in 
3: Multiply both sides by 27, resulting in 
4: Divide both sides by -8, resulting in 
5: Find the cube root of both sides, resulting in ![\sqrt[3]{-\frac{27t}{8} }=y](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-%5Cfrac%7B27t%7D%7B8%7D%20%7D%3Dy)
6: Apply a radical rule, resulting in ![-\sqrt[3]{\frac{27t}{8} } =y](https://tex.z-dn.net/?f=-%5Csqrt%5B3%5D%7B%5Cfrac%7B27t%7D%7B8%7D%20%7D%20%3Dy)
7: Apply another radical rule, resulting in ![-\frac{\sqrt[3]{27t} }{\sqrt[3]{8} } =y](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Csqrt%5B3%5D%7B27t%7D%20%7D%7B%5Csqrt%5B3%5D%7B8%7D%20%7D%20%3Dy)
8: Simplify the denominator, resulting in ![-\frac{\sqrt[3]{27t} }{2} =y](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Csqrt%5B3%5D%7B27t%7D%20%7D%7B2%7D%20%3Dy)
9: Apply yet another radical rule, resulting in ![-\frac{\sqrt[3]{27}\sqrt[3]{t} }{2} =y](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Csqrt%5B3%5D%7B27%7D%5Csqrt%5B3%5D%7Bt%7D%20%20%20%7D%7B2%7D%20%3Dy)
10: Simplify ![\sqrt[3]{27}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27%7D)
, resulting in ![-\frac{3\sqrt[3]{t} }{2} =y](https://tex.z-dn.net/?f=-%5Cfrac%7B3%5Csqrt%5B3%5D%7Bt%7D%20%20%20%7D%7B2%7D%20%3Dy)
DONT CLCIK THE LINK PLEASE REPORT IT
Check the picture below.
make sure your calculator is in Degree mode.
Answer:
,
Step-by-step explanation:
The function of the graph can be written in the vertex form as
, where V(h,k)=V(2,4) is the vertex of the quadratic function.
We substitute the value to obtain;
,
The point (5,1) lies on the graph so we use it to determine the value of a.
,
,
,
The required equation is
,
