# Find the x intercepts of the graph of y = -x^2 + 3x + 18

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### 3 Answers

We have the graph : y = -x^2 + 3x + 18

At the x-intercept y = 0

So to find the x- intercept we equate

-x^2 + 3x + 18 = 0

=> x^2 - 3x - 18 = 0

=> x^2 - 6x + 3x - 18 = 0

=> x(x - 6) + 3( x - 6) = 0

=> (x + 3)(x - 6) = 0

Therefore the x intercepts are:

**( -3 , 0) and ( 6, 0)**

y= -x^2 + 3x + 18

We need to find the x-intercept of the curve y.

The x-intercept is the point where the curve y meets the x-axis.

Then the values of y would be zero.

==> -x^2 + 3x + 18 = 0

==> We will factor -1.

==> -(x^2 -3x -18) = 0

Now we will factor.

==> -(x-6)(x+3) = 0

==> x = 6 , -3

There are 2 values for x. Then, the curve y meets the x-axis at two points ( 6,0) and (-3,0)

**Then x-intercepts are (6,0) and (-3,0)**

find x intercepts of y = -x^2 + 3x + 18.

The x intercepts of a graph are the points on x axis where the graph intersects. On x axis y coordinate is zero always.

To find the x intercept of y = -x^2+3x+18, we put y = 0 and solve for x.

0 = -x^2+3x+18.

We subtract -x^2+3x+18 from both sides:

=> x^2-3x-18 = 0.

=> (x+6)(x-3) = 0.

So x+6 = 0 or x-3 = 0.

**So ****the x intercepts of the graph are ****x = -6 or x = 3.**