# What is the closest point of the parabola y = 7x^2 + 9x + 3 to the x-axis?

*print*Print*list*Cite

### 2 Answers

Without using calculus, we know that the vertex is the maximum or minimum of the parabola. We can use the formula `x=(-b)/(2a)` for the x position of the vertex.

In this case a = 7, b = 9

This gives us the x coordinate of the vertex.

`x = (-b)/(2a)=(-9)/(2(7))=-9/14`

Now the y coordinate when `x=-9/14` we evaluate `y=7x^2+9x+3` at `x=-9/14`

We get

`y = 7(-9/14)^2 + 9(-9/14) + 3 = 7(81/196)+9(-9/14)+3=81/28-81/14+3 `

GCD = 28

`81/28-162/28+84/28=3/28`

So the closest `7x^2+9x+3` gets to the x axis is

`(-9/14, 3/28) ` Same answer as above but without Calculus.

The given parabola is y = 7x^2 + 9x + 3

From the equation of the graph we know that it opens upwards. The closest the graph of the parabola comes to the x-axis is when y is the least.

We have to determine the minimum value of y = 7x^2 + 9x + 3

This can be determined by solving y' = 0, for x

y' = 14x + 9

14x + 9 = 0

=> x = -9/14

y = 7*(81/14^2) + 9*(-9/14) + 3

=> y = 81/28 - 81/14 + 3

=> y = (-81 + 84)/28

=> y = 3/28

**The parabola comes closest to the x-axis at y = 3/28**