Answer:
x-intercepts at x = 1 and 2.5
y-intercept at y = 5
Step-by-step explanation:
Given: f(x) = 2x² - 7x + 5
It is required to find x and y intercepts.
x-intercept ⇒ the value of x when y = 0
y-intercept ⇒ the value of y when x = 0
So, to find x intercepts of the given function put f(x) = 0
∴ 2x² - 7x + 5 = 0
a = 2 , b = -7 and c = 5
Using the general rule to find the roots:
[tex]x=\frac{-b \pm \sqrt{b^2-4ac} }{2a} =\frac{-(-7) \pm \sqrt{(-7)^2-4*2*5} }{2*2} =\frac{7 \pm \sqrt{49-40} }{4}=\frac{7 \pm \sqrt{9} }{4}=\frac{7 \pm 3 }{4}[/tex]
∴ x = (7+3)/2 = 10/2 = 2.5
OR x = (7-3)/4 = 4/4 = 1
So, x-intercepts at x = 1 and 2.5
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To find y intercepts of the given function put x = 0
∴ y = 2x² - 7x + 5 = 2 * (0)² - 7 * 0 + 5 = 5
So, y-intercept at y = 5
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See the attached figure which represents x and y intercepts of the function f(x) using the graph.
