find the x and y intercepts:
f(x)=-x^2 +3x+10
for the y intercept let x = 0
for the x intercept, let y=0 or f(x) = 0
that is ...
solve x^2 + 3x + 10 = 0
Hint: there is no real solution, so there are not x intercepts
explanation:
You are looking at a parabola which opens upwards and whose vertex is above the x-axis, so how can the graph cut the x - axis ?
Thanks, I kept coming up with strange answers. That makes sense.
To find the x-intercepts of a function, we need to determine the values of x when f(x) equals zero. In other words, x-intercepts are the points on the graph where the function crosses the x-axis.
Let's set f(x) equal to zero and solve for x in the given function:
0 = -x^2 + 3x + 10
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. Let's use the quadratic formula in this case:
x = (-b ± √(b^2 - 4ac)) / (2a)
In the equation f(x) = -x^2 + 3x + 10, we have:
a = -1
b = 3
c = 10
Substituting the values into the quadratic formula:
x = (-(3) ± √((3)^2 - 4*(-1)*(10))) / (2*(-1))
Simplifying further:
x = (-3 ± √(9 + 40)) / (-2)
x = (-3 ± √(49)) / (-2)
x = (-3 ± 7) / (-2)
This gives us two possible solutions for x:
x₁ = (-3 + 7) / (-2) = 4 / (-2) = -2
x₂ = (-3 - 7) / (-2) = -10 / (-2) = 5
Therefore, the x-intercepts of the function f(x) = -x^2 + 3x + 10 are x = -2 and x = 5.
To find the y-intercept of a function, we need to calculate the value of f(x) when x equals zero. In other words, the y-intercept is the point on the graph where the function crosses the y-axis.
Let's substitute x = 0 into the given function to find the y-intercept:
f(0) = -(0)^2 + 3(0) + 10
f(0) = 0 + 0 + 10
f(0) = 10
Therefore, the y-intercept of the function f(x) = -x^2 + 3x + 10 is y = 10.