1. In the first line, type the (implicit) function: \( x+y=3 \) 2. In the second line, type the (explicit) function \( (t)=-*)+5 \) 3. Celcutate (by hand) the intercepts for \( x+y=3 \). \[ \begin{array}{cc} x+0=3 & o+y=3 \\ x=3 & y=3 \end{array} \] The \( y \)-intercept(s) of \( x+y=3 \) is/are: \( (0,3) \) The \( x \)-intercept(s) of \( x+y=3 \) is/are: \( (3,0) \) Once you find the intercept(s) algebraically, confirm your intercepts using Desmos. 4. Calculate (by hand) the intercepts for \( f(x)=-x^{2}+5 \). The \( y \)-intercept(s) of \( f(x) \) is/are: The \( x \)-intercept(s) of \( f(x) \) is/are: Once you find the intercept(s) algebraically, confirm your intercepts using Desmos. 5. Calculate (by hand) the point(s) of intersection of the two graphs.
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The Deep Dive
To find the intercepts for \( f(x)=-x^2+5 \), start with the \( y \)-intercept by evaluating \( f(0) \): \[ f(0) = -0^2 + 5 = 5. \] So the \( y \)-intercept is \( (0, 5) \). For \( x \)-intercepts, set \( f(x) = 0 \): \[ -x^2 + 5 = 0 \implies x^2 = 5 \implies x = \pm \sqrt{5}. \] Thus the \( x \)-intercepts are \( (\sqrt{5}, 0) \) and \( (-\sqrt{5}, 0) \). To find the intersection points of the two graphs, substitute \( y = 3 - x \) into \( f(x) = 3 - x \): \[ -x^2 + 5 = 3 - x \implies -x^2 + x + 2 = 0. \] Factoring gives \( (x-2)(x+1)=0 \), yielding \( x = 2 \) and \( x = -1 \). Substituting these \( x \)-values back to get \( y \)-coordinates: - For \( x = 2 \): \( y = 3 - 2 = 1 \) so \( (2, 1) \). - For \( x = -1 \): \( y = 3 - (-1) = 4 \) so \( (-1, 4) \). The points of intersection are \( (2, 1) \) and \( (-1, 4) \).
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